RADIOACTIVITY
Some elements are unstable. They spontaneously emit particles like alpha, beta or positron and
change to other elements. This phenomenon is called radioactivity. We know from the chapter
Periodic Table that all the isotopes of all the elements above atomic number 83(Bi) are radioactive
and are unstable. From 84-92, the elements are naturally radioactive and above 92 are synthetically
radioactive. However some isotopes of elements below atomic number 83(Bi) are also radioactive.
For example, C12 and C13 isotopes are stable while C14 isotope is radioactive; H1 and H2(deuterium)
are stable while H3(tritium) isotope is radioactive; K39 isotope is stable while K40 isotope is
radioactive and so on. Majority of the lighter isotopes are stable and are not radioactive. A few
of them are unstable. But above Bi, i.e from Po(84) onwards, no isotope is stable. The isotopes
are also called nuclides and so in this chapter we shall use the term nuclide in place of isotope
very often. The particles alpha, beta etc. emitted from the radioactive nuclide come out of the
nucleus of the nuclide. Alpha and beta are particles, not true rays which look like rays because
the partcles travel with vey high speed. Gamma is a electromagnetic radiaon i.e it is a true ray.
Let us know first about the different radioactive emissions.
ALPHA(α
α) PARTICLES:
An alpha particle is a particle consisting of 2 protons and 2 neutrons which come out of the
nucleus with high speed. The alpha particle is given a symbol 2He4 because it resembles with
dipositive He++ ion. A He++ ion contains 2 protons and 2 neutrons (i.e 2 electrons removed from
a He atom) like an alpha particle, hence the symbol. When alpha particle is lost by the nuclide,
the new nuclide called the daughter nuclide contains two protons less and so the new element
is placed two places left of the parent element in the periodic table.
92
U238 ----------> 2He4(alpha particle) + 90Th234
Note that the subscript gives the atomic number i.e the proton number while the superscript
gives the mass number(proton number + neutron number). When Uranium-238 isotope emits
one alpha particle, the daughter nuclide contains 90 protons and the new element is Th-234
isotope. Note that in nuclear reactions of the type shown above, the reader should see that the
mass number in the LHS should be same as the total mass number in the RHS(i.e the total of
superscripts of two sides should be equal), similarly the proton numbers of LHS should be same
as the total proton numbers of RHS(i.e the total of subscripts of two sides should be same).
238 = 4 + 234(mass number) and 92 = 2 + 90(proton number)
One thing is to be borne in mind that when one alpha particle is emitted by the nucleus of the
atom, the total positive charge of the atom decreases by two. But the number of electrons
remains the same(92 in the above case) and now electron number is more than the proton
number by two. So what happens to the electrical neutrality of the daughter nuclide? Does it
become dinegative charged ion? No, the electrical neutrality is maintained soon after the emission
of one alpha particle. The excess electrons are removed from the valence shell of the daughter
nuclide to the surrounding atmosphere to equalise the number of protons with number of
electrons(90 in this case). Hence the daughter nuclide becomes a neutral atom like the parent
element and not an anion.
1
More Examples:
Pu239 --------> ? + 2He4(alpha)
( ? = 92U235)
94
When Plutonium 239 isotope emits an alpha particle, the daughter nuclide is Uranium 235
isotope which is different from U-238 isotope taken in the first example.
Note that U-238 isotope is an alpha emitter(1st example) and Pu-239 isotope(2nd example)
is also an alpha emitter. Those nuclides which emit alpha particles are called alpha emitters.
They cannot be beta emitters, i.e those isotopes cannot emit beta particles.
Another important thing you remember that alpha particle is emitted only by heavier
nuclides i.e usually nuclides above atomic number 82(Pb), and not by any ligher nuclide
such as 6C14 or 19K40 etc.
SAQ 1: The following nuclides are alpha emitters. Predict the daughter nuclides and write the
nuclear equation when each nuclide undergoes alpha decay.
(i)90Th230
(ii)90Th232
(iii)88Ra226
(iv) 86Rn220
BETA(β
β) PARTICLES:
A beta particle is identical with an electron. It has negligible mass(nearly 0) and charge equal to
-1. It is given the symbol -1e0. Always the subscript represents the charge and superscript
represents the mass. The question is how a beta particle which is nothing but an electron is
produced in the nucleus. A beta particle is not the electron already present in the atom in the
extra nuclear part. It is a new particle looking like an electron in mass and charge produced at
the nucleus and is expelled out from it at a tremendous speed. They are also wrongly called
β-rays.
A neutron in the nucleus is transformed to a proton and an electron(beta particle).
0
n1 --------> +1p1 + -1e0
(beta particle)
Thus when this beta particle is driven out of the nucleus, the nucleus now contains one proton
excess than the parent nuclide, although the mass remains nearly same. So the daughter element
contains one excess proton and occupy one place right in the periodic table. Look to this example.
90
Th234 --------> 91Pa
234
+ -1e0 (beta particle)
When Th-234 isotope emits a beta particle, the daughter nuclide occupies one place right of it,
that is Pa-234(protactinium) isotope is formed. Note that the parent nuclide had 90 electrons in
the extranuclear part(90Th232) but now the daughter has 91 protons with same number(90) of
electrons. What will happen to the electrical neutrality of the daughter nuclide? It becomes
electrically neutral by absorbing one electron from surrounding atmosphere and equalising the
number of protons with number of electrons. Remember that atmosphere is a great reservoir of
electrons. It can donate and accept electrons whenever required.
More example:
C14 --------> ? + -1e0 (beta)
(? = 7N14)
6
Po218 -------> ? + -1e0 (beta)
(? = 83Bi218)
82
Note that the nuclide which is a beta emitter like C-14 isotope and Po-218 isotope are only
2
beta emitters. They can never be alpha emitters.
Unlike alpha emission, which takes place only by heavier nuclide(>Pb), beta emission
takes place by both lighter nuclide( eg. C-14) and heavier nuclide(Po-218).
SAQ 2:The following nuclides are beta emitters. Write the nuclear equations and predict the
daughter nuclides.
(i) 90Th234
(ii) 19K40
(iii)91Pa234
(iv)27Co60
212
(v) 83Bi
SAQ 3: 92U238 isotope is a ______emitter and its daughter nuclide after alpha emission is a
_____emitter. Is there any relationship between the nature of emission of the parent and daughter
nuclides?
POSITRON EMISSION:
Positron is another particle which is emitted from certain radioactive nuclides. Positron is not a
fundamental particle like electron or proton. It is only produced during radioactive decay of
certain radioactive nuclides. Positron is analogous to an electron in mass(negligible mass) but
also analogous to a proton in charge(+1 charge). It is given the symbol +1e0. Now the question is
how a positron is produced in the nucleus. A proton is transformed to a neutron and a positron.
Look to the following equation.
p1 --------> 0n1 + +1e0 (positron)
+1
After the positron is driven out, the daughter nuclide has proton number less than that of the
parent nuclide by one, although mass remains nearly constant. Thus the daughter nuclide is a
new element occupying one place to the left of the periodic table. Let us look to this example.
Na22 --------> +1e0 + 10Ne22
11
Al25----------> +1e0 + 12Mg25
13
Note that Na-22 and Al-25 isotopes are positron emitters? The normal Na and Al that we
know are Na-23 and Al-27 isotopes which are stable and not radioactive. But Na-22 and
Al-25 isotopes which are prepared artificially are radioactive and are positron emitters. When
they expel a positron from their nuclei, they are converted to new elements Ne-22 and Mg-25
isotopes respectively. So in general when a positron is emitted the new daughter element
occupies one place left of it in the periodic table. After the loss of a positron, the nucleus now
has one proton short compared to the parent but the number of electrons remains the same. The
electrical neutrality of the daughter species is maintained by the immediate loss of one electron
from the valence shell of the daughter to the surrounding atmosphere so that the number of
protons and electrons remain the same in the daughter nuclide. Thus the daughter is electrically
neutral and not a negative charged ion.
Note that the nuclide which emits positron is a positron emitter. It cannot be either alpha or
the beta emitter.
Only lighter nuclides(<<<Pb) emit positrons and not the heavier nuclides(>Pb).
GROUP DISPLACEMENT LAW:
We already know that when an alpha particle is emitted by any radioactive nuclide, the daughter
element occupies two places left of it in the periodic table, when a beta particle is emitted, the
3
daughter element occupies one place to the right and when positron is emitted the daughter
element occupies one place to the left of the periodic table. This is called the Group displacement
law developed by Soddy.
GAMMA (γ) RAYS:
Gamma rays are real rays and not particles. They are a form of electromagnetic radiations
having small wavelengths (high energy). You have studied about it in the chapter atomic structure.
It is given the symbol 0γ0 i.e it carries neither any mass nor any charge. Gamma rays are formed
in two ways.
(1)
When a radioactive nuclide expels a alpha or a beta particle in very high speed, it is
often excited to a higher energy level. Imagine when you vomit out food from your stomach,
how do you feel? Are you not excited? In the same manner when the nucleus vomits out an
alpha or a beta particle in a tremendous speed, it is excited to a higher energy level. After
some time, it comes back to the ground state by emitting one photon of a short wavelength
radiation called gamma(γ) radiation. Do you know that the gamma radiations that come out after
the beta emission of a Co-60 isotope is used in the treatment of cancer? This is called radiotherapy
for cancer.
Co60 -------> -1e0 + 28Ni60** (2nd excited state)
γ photon
Ni60** -------> 28Ni60* (1st excited state) +
28
60
60
Ni * --------> 28Ni (ground state) + γ photon
28
Thus after the emission of a beta particle by a Co-27 isotope, the new nuclide Ni-60 climbs up to
the 2nd excited state. It comes back to ground state in two successive jumps and thereby emit
two photons of gamma rays.
To conclude, gamma is not a particle like alpha, beta and positron. It is an energy radiation which
is released after the emission of alpha, beta or positron. An alpha emitter can emit gamma ray
photon, a beta emitter can emit a gamma ray photon. Note that, not all radioactive nuclides emit
gamma ray photon after the emission of particles like alpha, beta. There are some which do so,
not all. For example, C-14 isotope does not emit a gamma photon after emitting a beta particle,
while Na-24(artificial) isotope emits a gamma photon after emitting a beta particle.
(2)
The other way by which a gamma photon is formed is positron-electron annihilation.
This happens in case of positron emitters in isotopes having lower n/p ratio than expected. In
some cases, positron instead of going out of the atom unites with an electron of the valence shell
of an atom. When positron unites with an electron, both of them are destroyed in a process
called annihilation and the mass is coverted to energy. The positron-electron annihiliation produces
two gamma photons moving in opposite directions.
27
SAQ 4: Write the radioactive decay reaction for the following nuclide. The particle emitted in
each case has been mentioned within bracket. Refer a periodic table to locate the position of the
daughter nuclide.
(i)12Mg27(β)
(ii)83Bi210(β)
(iii)84Po213(α) (iv) 8O15(positron)
200
82
(v)79Au (β)
(vi)35Br (β)
(vii)88Ra226(α) (viii)84Po210(α)
38
122
(ix)19K (positron)
(x)53I (positron)
(xi)26Fe60(β), (xii)93Np230(α)
(xiii)23V46(positron)
(xiv)87Fr221(α)
4
PROPERTIES OF ALPHA, BETA, POSITRON AND GAMMA RAYS:
ALPHA:
These are like He++ ions consisting of two protons and two neutrons. They are ejected out from
the nucleus with high velocities in the range of 1.7 X 107 cm/sec which is 6% of the velocity of
light. They are deflected to the negative plate of an electric field. The deviation towards the
negative potential is smaller compared to beta particles as they possess higher kinetic energy.
They can penetrate through thin metal sheets. Do you remember the Rutherford's gold foil
experiment? Alpha particles penetrated through a thin gold foil of 0.00004cm thickness. These
have the least penetrating power and can be stopped by aluminium sheet of 0.1mm thick. It
cannot penetrate metal sheets above this thickness. Alpha particles cause ionisation of the gas
through which they pass. Its ionising power is the maximum because it contains the highest
kinetic energy. They affect the photographic plate. When alpha particles strike on a ZnS screen,
produces luminescence(ZnS glows).
BETA:
These are nothing but electrons( -1e0) having -1 charge and negligible mass. They travel with
high velocity often higher than alpha particles(nearly equal to velocity of light- 99% of velocity of
light). They are deflected to the positive plate of an electric field and the extent of deviation is
more as they posses less kinetic energy than alpha particles. They have more penetrating power
than alpha particles. They can penetrate through thicker metal sheets than alpha particles.They
can be stopped by aluminium sheet of 1cm thick. Since they possess lower kinetic energy(due to
negligible mass, E = mc2), they ionise gases to a smaller extent than alpha particles. They also
affect photographic plate to a greater extent and also produce glow in ZnS screen but to a lesser
extent than alpha particles.
POSITRON:
These carry +1 charge like protons and negligible mass like electrons and are represented as
e0.Therefore they are sometimes called as positive electrons. They travel with high speed like
+1
beta particles and are deflected towards the negative plate of the electric field. Other properties
are analogous to beta particles.
GAMMA RAYS:
These are not particles and are high energy(low wavelength) electromagnetic radiations. They
do not carry any charge or mass. They travel with the velocity of light. They are not deflected by
electric or magnetic field. They have the highest penetrating power, they can pass through much
thicker aluminium sheets. They can be stopped by aluminium sheet of 100cm thickness. They
ionise the gases but to the smallest extent as they do not carry any mass. They affect to the
smallest extent to the photographic plate and also produce glow in ZnS screen.
TYPES OF RADIOACTIVITY:
There are two types of radioactivity, i.e
(i)Natural radioactivity:
Those nuclides which are present in the nature and
are radioactive are called naturally radioactive. The phenomenon is called natural radioactivity.
All the isotopes of all elements beyond atomic number 83(Bi) are naturally radioactive. However
some lighter isotopes such as 6C14, 19K40, 43Tc99 etc. are also naturally radioactive.
(ii) Artificial Radioactivity: Those radioactive nuclides which are artifically prepared
give rise to the phenomenon of artificial radioactivity. Majority of the lighter nuclides that are
5
radioactive such as 5B10, 11Na22, 27Co60 etc. are all artificially prepared and are radioactive.
THEORY OF RADIOACTIVITY:
It is very surprising to believe that although protons are positively charged particles, remain
together in a small space called nucleus without repelling each other. What factor is responsible
for binding the protons together? It is the neutral particles- neutrons which remain in between
protons and are responsible to bind the protons in the nucleus. Neutrons and protons are
jointly called nucleons. The neutrons act like cement to bind the protons together, so that we get
a stable nucleus. For very lighter nuclides i.e upto atomic number 20(Ca), when the proton
number is small and the repulsion between the protons is not high, the number of neutrons
necessary to make stable nuclei is nearly equal to the number of protons present. For example,
C12, 8O16, 20Ca40 etc. in which the number of protons is equal to the number of neutrons.
6
Although many exceptions are there like 5B11, 11Na23, 13Al27 etc. in which the number of neutrons
is greater than number of protons by one, the nuclides are still stable. Such deviations are not
considered abnormal, as the law of nature at times does not follow a fixed order or rule as
predicted by human beings and it is not possible to explain its cause. Anyway, upto atomic
number of 20(Ca), the neutron to proton ratio(n/p) is nearly equal to one. As the atomic
number increases beyond 20, the repulsion between the protons become more and more severe.
To overcome that more and more number of neutrons are necessary to make stable nuclides. So
the neutron to proton ratio(n/p) steadily increases for the stable nuclides as we move on to
higher atomic numbers. Let us take the examples of 26Fe56, 35Br79, 80Hg200 which are stable
nuclides. The number of neutrons in them respectively are 30, 44 and 120. If we take the n/p
ratio of these three nuclides, they are 30/26=1.153 for Fe, 44/35=1.257 for Br and 120/80=1.5
for Hg. So, what do we find here? For getting stable nuclide, the neutron number should be more
and more i.e the n/p ratio has to increase gradually as the proton number increases beyond 20. It
should not remain close to 1. This is required to overcome the repulsive forces between protons
which become more and more severe. Speaking in a different way, we can say that Fe nucleus
would have been unstable if n/p ratio would have been 1 instead of 1.153 i.e if it would have 26
neutrons in stead of 30. Similarly Br and Hg nuclei would have been unstable if they would have
neutrons less than 44 and 120 respectively. Upto atomic number 83(Bi), we find at least one
isotope of every element(except Tc and Pm) which are stable. The n/p ratio gradually increases
from nearly 1 at Ca(20) to nearly 1.5 at Bi(83). But look to the law of nature after atomic
number 83!!! For elements beyond Bi(83), the repulsive forces become so severe that, no
excess number of neutrons is able to bind the protons and make a stable nucleus. So all the
isotopes of all the elements beyond Bi(83) are unstable nuclides and are spontaneously radioactive.
They emit alpha or beta particles and change to another element. The whole purpose behind this
emission and transformation is to acquire more and more stability in the daughter nuclei. The
emission of radioactive particles (a and β) continues in sequence by a parent to give daughter,
then daughter giving grand daughter and grand daughter giving great grand daughter and so on
until the last nuclide acquires atomic number 83 or less. Usually the nuclides ultimately transform
to one of the isotopes of lead(Pb-82). Look to this sequence of natural radioactivity.
92
U238 -----α ------> 90Th234 ------β------>91Pa234 ------β-----> 92U234 ------α---->90Th230
6
------α-----> 88 Ra 226 ------α-------> 86 Rn 222 ------α------> 84 Po 218 ------β----> 85 At 218
------α----->83Bi214 -----α------> 81Tl210-----β----->82Pb 210-----β------>83Bi210----α----> 81Tl206
------β----> 82Pb206(Stable)
U238 isotope is an alpha emitter and its daughter 90Th234
is a beta emitter and its daughter 91Pa234 is also a beta
emitter. In this way, every daughter produced from its
parent is unstable and decays either by emitting an alpha
or a beta particle. At last when a stable isotope is formed,
radioactivity is stopped. 82Pb206 isotope is the end nuclide
in the above series. This is called Uranium(238) series.
The stability of different nuclides is best known from
the graph of neutron versus proton for different nuclides
as shown in the following diagram called the Band of
Stablity.
The 450 straight line is the n/p=1 line, that means if any
nuclide will fall on this line, its neutron and proton
numbers are equal. Upto proton number 20(Ca), n/p is
nearly 1, and nuclides fall more or less near the 450 line
but beyond Ca the n/p ratio gradually increases as explained before. The dotted band in the
graph is called the band of stability. Because the n/p ratio of stable nuclides above Ca increases
steadily and so the stability zone, instead of going straight, rises up in a curved path whose slope
gradually increases after calcium. Note that, in stead of a curved line, there is curved band as
always there are exceptions and deviations to the general rule that has been theoretically
formulated. Any nuclide which falls inside the stability band is a stable nuclide and not radioactive.
The stability band is shown upto atomic number 83. That means upto atomic number 83, we get
stable nuclides, although some nuclides within this region too are unstable. But above 83(Bi),
there is no isotope of any element which is stable and hence all are radioactive. Some of them
emit alpha and some beta particles and the emission successively continues until the last nuclide
falls within the stability band(<83). There is no hard and fast rule for the nuclides having atomic
number above 83, the way each will decay. Some emit alpha particles and some beta particles
and there is no logic for this. But all the natural radioactive nuclides fall into four different series
which will be discussed later. But there is always a rule and reason for the decay of the radioactive
nuclide which lie below 83. Let us know about this.
92
Any nuclide which falls outside the band of stability is unstable and hence radioactive. For
example, 11Na24 isotope does not fall in the band of stability. The number of neutron=13 and the
number of protons=11 and n/p=1.2(nearly), it has more number of neutrons than what is required
for its stable nucleus(11Na23). So the nuclide is unstable and radioactive. Thus 11Na24 isotope will
fall above the band of stability. We know that 11Na23 isotope is the stable isotope of sodium
although it has 12 neutrons and 11 protons and n/p ratio is slightly greater than one(exceptions
are always there).Take the case of 11Na22. For this nuclide, proton and neutron number are
same.Although theoretically it should be a stable nuclide(n/p=1), actually it is not stable nuclide
and is radioactive. While preparing the band of stability, actual stability of all the nuclides
7
were taken into account, not theoretically drawn on the basis of any strict rule. The reader
should not think that n/p ratio will be always exactly equal to 1 for stable nuclides upto atomic
number 20. It may be, in some case, equal to 1, and in some other cases slightly greater than
1(say in case of 11Na23, 5B11 etc. which are stable nuclides) . What is more important is to rely
on the experimental stability band. Let us remember now the common stable isotopes of some
common elements.
STABLE ISOTOPES OF COMMON ELEMENTS:
Some of the common elements and their stable isotopes are as follows:
Be9,
B11, 6C12 and 6C13, 7N14, 8O16, 8O17, 8O18, 9F19, 10Ne20, 11Na23, 12Mg24, 13Al27, 14Si28,
4
5
29
31
Si , 15P , 16S32, 16S34, 17Cl35, 17Cl37, 18Ar40,
K 39, 20Ca 40, 35Br 79, 26Fe 56, 27Co 59, 35Br 81,
14
19
108
Ag and so on. Excepting a few, in all these cases the n/p ratio is not exactly 1. Note that
47
Be-9 isotope is stable and falls inside the band of stability but Be-10 isotope is unstable and falls
outside and in this case above the band of stability (since n/p ratio is greater than expected). B11 is stable and falls inside the band of stability while but B-10 is unstable and falls outside the
band and in this case below the band of stability(as n/p ratio is less than expected). C-12 and C13
are stable nuclides and fall inside the band while C-14 falls above it. Na-23 is stable but Na-24
and Na-22 are unstable. Na-24 falls above the band while Na-22 falls below the band.
SAQ 5: Indicate whether the following nuclides fall within, or above or below the band of
stability. Which ones are unstable nuclides and hence radioactive?
Mg-27, P-30, Cl-37, K-39, K-40, O-15, Ne-20, C-13, B-10, Br-82, Fe-60, Ag-111
BETA and POSITRON EMISSION FOR LIGHTER NUCLIDES:
It has been already told before that for heavier nuclides above atomic number 82, there is no
fixed rule as to which nuclide will emit alpha particle and which will emit beta particle. However
there is a fixed rule for emission in case of lighter nuclides.
Case-I: Unstable nuclides which lie above band of stability(Beta emission):
The nuclide which lies above the band of stability(n/p ratio is greater than expected) i.e the mass
number is more than what is expected of its stable nuclide, is a beta emitter. By emitting beta
particle, it tends to stabilise itself. You know that a beta particle is formed by the conversion of a
neutron into a proton and a beta particle. After the loss of beta particle, the nucleus gains one
proton and loses a neutron. Thus the n/p ratio decreases. The new daughter nuclide usually
becomes a stable nuclide and falls inside the stability band and does not further decay. Look to
the following case.
Mg27 ---------> -1e0 + 13Al27
12
Mg-27 isotope is unstable and radioactive as it has more number of neutrons than expected of its
stable nuclide(12Mg24). It is transformed to the stable 13Al27 isotope by emitting a beta particle.
So remember that any nuclide whose mass number appears to be more than what is
commonly known is radioactive and a beta emitter. Look to another example.
C14 ---------> -1e0 + 7N14
6
C-14 lies above the band of stability and hence is radioactive. It decays by beta emission, so that
it is converted to a stable N-14 nuclide. Solve the following SAQs.
8
SAQ 6: Why the following are beta emitters? Give the daughter nuclides formed in each case.
H3, 35Br82, 26Fe60, 27Co60, 47Ag111
1
Case-II:
Unstable nuclides which lie below the band of stability(positron
emission):
The nuclides which lie below the band of stability(n/p ratio less than expected) i.e mass numbers
are less than those of their stable nuclides, are positron emitters. By emitting a positron, it tends
to stabilise itself. You know that a positron particle is formed by the conversion of a proton into
a neutron and a positron particle. After the loss of positron, the nucleus gains one neutron and
loses a proton. Thus n/p ratio increases and the daughter nuclide tends to become more stable.
Look to this example.
8
O15 ---------> +1e0(positron) + 7N15,
19
K38 -------->
e0 + 18Ar38
+1
In these cases the n/p ratio of each daughter element is more than that of its parent nuclide.
Calculate for yourself to verify this. Both the daughter nuclides formed in the above examples
are more stable than their parents.
SAQ 7: Explain why the following are positron emitters? Give the daughter nuclide formed in
each case.
P28, 7N12, 10Ne18, 11Na21, 21Sc41, 16S30, 23V46, 25Mn50
15
SAQ 8: Which of the following is a beta emitter and which a positron emitter and why?
C14, 5B10, 11Na22, 11Na24, 19K40, 19K37, 12Mg27, 25Mn56, 53I129, 15P28, ,21Sc41, 27Co60
6
The stable nuclides have the mass numbers as follows: C-12, B-11, Na-23, K-39, Mg-24, Mn-55,
I-127, P-31, Sc-45, Co-59,
HALF LIFE PERIOD
All radioactive elements decay in a very interesting style. The process of their disintegrtion is
called a first order reaction. The nature of a first order reaction is that the time required to
complete half of the reaction is always same. For example, let us take 100g[A0]m of a radioactive
substance (say 92U238). Let the time taken for the disintegration of half of it is T1 years. So after
T1 years, we find that 50gms have disintegrated to its daughter and remaining 50gms of U[A0/2]
would be present. Now guess what time is
required to complete half of 50gms which is
present now. It will take another T1 years.
That means for disintegrating half of 50gms
i.e 25 gms it will take the same time(T1 years)
as it took for disintegrating 100gms to 50gms.
So after two half life periods i.e 2T1 years,
the amount of 92U23 left is 25gms[A0/4].
Now what time is required for disintegrating
half of this value? Another T1 years are
required for disintegrating half of 25gms i.e
12.5 gms and leave back 12.5gms of the
parent(U238)nuclide. So after 3 half life
periods(3T 1), the amount left will be
9
12.5gms[A0/8] .This trend continues until all the substance is exhausted. Truly speaking, it will
take infinite time for the completion of the reaction, i.e to totally exhaust the parent nuclide, but
practically the reaction is nearly 99.9% complete after 10 half life periods. This type of graph
is called exponential decay graph. The disintegration of radioactive elements and the concept of
half life periods are depicted in the given diagram. After nearly 5 half life periods, the graph
comes close to x-axis which indicates that the process is nearly going to be completed.
Very interestingly the half life periods of radioactive elements vary from few seconds to several
years. It varies from radioactive nuclide to radioactive nuclide. The half life periods of some
common radioactive nuclides are given below.
Half Life periods of Some Naturally Radioactive Elements:
NUCLIDE
Uranium(U-238)
Radium(Ra-226)
Radium(Ra-228)
Thorium(Th-230)
Radon(Rn-222)
Carbon(C-14)
HALF-LIFE
4.5 X 109 yrs
1590yrs
6.7yrs
8000yrs
3.82days
5,730 yrs
EMISSIONS
alpha
alpha
beta
alpha
alpha
beta
Half Life periods of Some Artificially Radioactive elements
NUCLIDE
Tritium(H-3)
Oxygen (O-15)
Sodium(Na-24)
Phosphorus(P-32)
Iodine(I-131)
HALF-LIFE
12.26 yrs
123 secs
14.96 hrs
14.28 days
8 days
EMISSIONS
Beta
Positron
Beta
Beta
Beta
We find that for 92U238 isotope the half life is so high i.e 4.5 X 109 years, while for 86Rn222 it is
few days(3.82 days ) while for 8O15 it is 123secs. The element which has a high half life will take
a long time to exhaust while those having small half life will end soon. That is why many naturally
radioactive elements (Tc, Pm) which have very short half life periods are almost absent now.
Such elements have almost completely died due to their small half lives.
There is a short and handy formula which you can use at this stage to find out the amount of a
radioactive element left after a particular time period of observation. For that you have to know
the half life period of the element.
A t = A0 1
2
t
T0 .5
Where At = amount of substance left, A0=amount of substance initially taken,
t= timeof observation,
T0.5= half life period , t/T0.5 = number of half lives
10
Example: O-15 is a positron emitter having half life of 123secs. How much of O-15 will
remain after 10minutes and 15 secs if we take initially 2gms of it?
Solution:
Time =t = 10X60 + 15 = 615 secs.
At = 2(1/2)615/123 = 2(1/2)5 = 2/32 = 1/16gm.
So the amount left after 10minutes 15 secs will be 1/16gm.
SAQ 9:
(i)
C-14 isotope is a beta emitter and has a half life of 5730 years. If you take 10gms of
C-14, how long it will take to be reduced to 2.5 gms?
(ii)
You have 1 gm of 92U238 isotope. How long will it take to reduce to 1/16gm? Half life
period of U-238 is 4.5 X 109 years.
(iii)
How much of tritium will be left from 12.8gms of tritium after 7 half life periods? The
half life period of tritium s 12.26 years.
CARBON DATING
By this method we can know the age of an old object from plant or animal containing carbon. For
example, there was excavation (digging the earth) in the Mahenzodaro and Harrappa. During
excavation, an old wooden bowl was found. By carbon dating method, it was possible to know
the age of this bowl i.e when this wooden bowl was prepared(i.e when the tree from which the
bowl was prepared was cut). So indirectly we can know the approximate period before which
the Mahenzodaro and Harappan civilization existed.
We already know that C-14 isotope is radioactive and is a beta emitter. C-14 decays slowly as its
half life period is large i.e 5730 years.
C14 ---------> 7N14 + –1e0
6
There are two types carbon dioxide in our atmosphere, i.e C12O2(ordinary) and a very small
amount of C14O2(radioactive). C-14 isotope is formed in the atmosphere by the bombardment of
neutrons coming from sun(cosmic neutrons) with the ordinary nitrogen(N-14).
N14 + 0n1 -------> 1H1 + 6C14
7
A N-14 atom gets converted to a hydrogen atom and C-14 isotope, when a cosmic neutron hits
an ordinary atmospheric nitrogen atom. This C-14 atom reacts with oxygen gas to give
C 14 O 2 (radioactive carbon dioxide). Plants during photosynthesis take up this
C14O2(radioactive)along with the normal C12O2. So C-14 remains present in very small quantities
in the carbohydrate that plant prepares during photosynthesis. As long as the plant is living, the
C-14 content in the plant remains fixed, since photosynthesis occurs everyday during the day
time. Although C-14 is not stable and it is disintegrating by emitting a beta particle, its quantity
will not decrease in a living plant because of continued photosynthesis. But as soon as a plant
dies, it cannot perform photosynthesis and C-14 content starts to decrease with time as it decays.
But we know that the decay of C-14 is very slow as its half life period is 5730 years. So even
after thousands of years, the wooden bowl made from a dead plant still would possess radioactivity.
Note that the radioactivity of a material is often measured in terms of the number of
particles(beta in this case)emitted by one gm of substance per minute. This is done by an
instrument called Geiger Muller Counter. In a living plant, whether living now or living thousands
of years before(say Mahenzodaro or Harappan period), the radioactivity is fixed(as already
mentioned before). It is nearly 15.3 beta particles per minute per gram of the material. This
11
means that whether belonging to our time now or during Mahenzodaro's time, one gram of a
living plant emits 15.3 beta particle per minute(which means 153 beta particles in 10 minutes).
If we measure the beta particle activity in case of the old wooden bowl that we have got during
the excavation, we will find that the activity is less than 15.3. Depending on the time passed
since the cutting of the tree from which the bowl had been made, the activity of the old object
will vary. Suppose, the old wooden bowl emits 9 beta particles per minute per gram now(which
we can measure from the Geiger Muller Counter), we can find the age of the bowl (i.e the time
when the tree from which the bowl was made died) by comparing the activity of the old bowl
with activity of a living plant at that time. We already know that there is no difference between
the activity of a living plant now(=a living plant then). By knowing the age of the bowl, we can
indirectly know the time when that civilisation existed. In fact, by carbon dating method, the
scientists could know the time when old civilizations grew in different parts of the world. For
calculating the age of the bowl we shall use the same equation we have used earlier for the
radioactivity. The detailed mathematical aspects of these equations will be covered in higher
classes. For now, this simple method of calculation will solve the purpose of understanding
the concepts.
At = A0 1
2
t
T0 .5
Let us take Rt= radioacvitity after time t and R0 =radio activity of a living plant
Rt = R0 1
2
t
T0 .5
⇒
1
9 = 15.3
2
t
T0 .5
We know that T0.5(half life period) of C-14 is 5730 years. By substituting the value of T0.5 in the
above equation, we can solve the equation and find the age of the bowl(t). If you have studied
logarithm in mathematics by now, it is very simple to solve this equation by taking log on both
the sides. If you have not yet learnt logarithm, then wait until you learn it. After that you take up
the calculation part of carbon dating. Now only understand the concepts clearly.
1
log 9 = log 15.3 + log
2
t
5730
⇒ 0.9542 = 1.1847 +
t log 1
2
5730
By solving this equation, we get, t = 4387.92 years.
So the age of the wooden bowl recovered from the excavation is 4387.92 years. In other words
the civilization existed nearly 4387.92 years before. In this way, the carbon dating is helpful in
finding not only the age of an old carbon containing material such as wooden bowl, bone etc.
found in an archeological excavation but also knowing the period when an old and perished
civilisation grew. Take for example, bones of the dinosaurs species those existed thousands of
years ago could be found in many areas of South Africa and American jungles during excavation.
The carbon dating method could reveal the period when the dinosaurs species existed. You
know that dinosaur does not exist now and many English films have been made on this animal,
dinosaur in which the animal is prepared artificially on the basis of the bones found during
excavation.
12
SAQ 10: A dog femur bone is found during an excavation. The age of the excavated layers is of
archaeological interest, so dating by radiocarbon was carried out. The half-life of carbon-14 is 5
730 years. The beta activity of a living carbon-containing matter(living dog) is 15.3 beta particles
per minute per gram. The dog femur bone found, showed an activity of 12 beta particle per
minute per gram. Calculate the age of the bone in years.
MASS DEFECT AND BINDING ENERGY:
When a nucleus is formed, some mass is lost and is converted to its equivalent amount of energy.
From the Einstein equation, we can know the relationship between mass loss and energy obtained.
E = (Δ
Δm) c2
(where Δm is the mass loss and c is the velocity of light and E is the energy obtained).
This energy released when the nucleus is formed is called its binding energy of the nucleus.
Let us take the simple example of a 2He4 atom. How much of energy is released when helium
nucleus is formed. We know that the nucleus of He contains two protons and two neutrons. The
masses of each proton, neutron and electron are known to us in amu scale. Let us find
theoretically the atomic mass of He atom.
Mass of one proton = 1.0073 amu, Mass of one neutron = 1.0087 amu,
Mass of one electron=0.00055amu
The atomic mass of He= 2 X 1.0073 + 2X 1.0087 + 2 X 0.00055 = 4.0331 amu(theoretical mass)
But actual atomic mass(isotopic mass) of He as obtained by mass spectrometer(please refer the
chapter atomic mass for this) is 4.0015 amu.
Calculated atomic mass = 4.0331amu,
Actual isotopic mass = 4.0015amu
Do you find a difference between the calculated (theoretical) mass and actual mass of He
atom? Where did this differential mass go? This difference is called Mass Defect. This mass
has been lost from the nucleus in the form of energy during its formation. This is called the
binding energy of the nucleus. We can calculate this by using the Einstein's equation(E = Δmc2)
Δ M = 4.0331 - 4.0015 = 0.0316 amu
We know in the chapter, mole concept that 1 amu = 1.67 X 10-24 gm
So Ε = Δm c2 = 0.0316 X 1.67 X 10-24 gm X (3 X 1010 cm/s)2 = 4.74 X 10-5 erg
This is the binding energy of He nucleus.Binding energy is often expressed in the unit MeV(million
electron volt).
1 eV = 1.602 X 10-19 joule =1.602 X 10-12 erg.
So 1 MeV = 1.602 X 10-12 X 106 = 1.602 X 10-6 erg
So the energy corresponding to the mass defect of 1amu =
1X 1.67 X 10-24gm X (2.988 X 1010 cm/sec)2 = 14.91 X 10-4
erg.
Converting the energy from erg to MeV,
14.91 X 10-4 erg = (14.91 X 10-4)/(1.602X10-6) ≈ 931.6MeV
So 1 amu ≈ 931.6 MeV
We can use this relationship directly to convert mass defect(difference in calculated mass and
actual mass) to binding energy in MeV. In case of He atom, we found the mass defect,
Δm = 0.0316amu.
So the binding energy = 0.0316 X 931 = 29.42 MeV
13
BINDING ENERGY PER NUCLEON:
For comparing the stability of different nuclides, binding energy per nucleon is found out. It is
obtained when the binding energy of a nuclide is divide by the number of
nucleons(protons+neutrons) present in its nucleus.
Binding Energy per nucleon = (Binding Energy)/n
(where n= number of neutron + no. of protons)
In case of He nuclide, we have just discussed, the binding energy per nucleon = 29.42/4 =7.35
MeV
SAQ 11: Find the binding energy(B.E) and also binding energy per nucleon(B.E.per nucleon)
for
(i)7N14 (isotopic mass = 14.0067amu)
(ii) 15P31(isotopic mass = 30.9740amu)
Mass of one proton = 1.0073 amu, Mass of one neutron = 1.0087 amu, Mass of one
electron=0.00055amu
Between the two which has a higher binding energy per nucleon.
BINDING ENERGY PER NUCLEON - MASS NUMBER GRAPH :
Binding energy per nucleon versus mass numbers of all natural elements are given in the above
diagram. BE/N increases with increase in mass number drastically upto Fe(56) and then decreases
graudally with further increase in mass number. Between mass number 40 to 140( Cr, Mn, Fe,
Co, Ni, Cu etc) the BE/N values ranges between 8.2 to 8.7 MeV with an average of 8.4 MeV.
These constitute the nuclides having intermediate mass numbers which have the highest average
BE/N. These nuclides have greater BE/N than lighter nuclides below mass number 40( H, He,
C, N, O etc) as well as than heavier nuclides above mass number 140(Pb, Bi, U, Th etc.). For
example 92U235 has has BE/N of 7.5 MeV while 2He4 has BE/N of 7.0MeV and 26Fe56 has BE/
N of 8.7 MeV.
14
NUCLEAR ENERGY DURING RADIOACTIVE DECAY:
In the previous section, we knew that during the formation of a nucleus there is loss of mass(mass
defect) and this mass lost is converted into energy. We shall now see that mass loss also occurs
during a radioactive disintegration process. In other words when any radioactive nuclide emits a
particle like α, β or a positron, there is always a net loss of mass which is converted to its
equivalent amount of energy. The tremendous amount of kinetic energy which the ejected
particle possesses is due this energy. Let us take an example.
α -emission:
When the radioactive nuclide 84Po210 emits an alpha particle, it is converted to 82Pb206 nulclide.
We know the actual isotopic masses (experimentally determined) of each of the parent and
daughter nuclide and also the mass of an alpha particle.
Po210 ---------->
He4 +
Pb 206 + K.E of α + 0γ0
84
2
82
209.9829amu
4.0026amu
205.9745amu
The masses given below each species in the above equation are the actual isotopic masses of
the nuclides obtained experimentally. We can find the total mass of RHS and find the difference
in mass between LHS and RHS.
Mass of RHS = 4.0026+ 205.9745 = 209.9771 amu
Mass of LHS = 209.9829, so the loss in mass, Δm = 209.9829 - 209.9771 = 0.0058 amu
This amount of mass has been lost when 84Po210 nuclide disintegrates to 82Pb206 by emitting an
alpha particle. This mass has been converted to energy as per Einstein's equation. We know that
for a mass loss of 1 amu the energy obtained is 931.6MeV, so the energy evolved due to mass
loss of 0.0058amu = 931 X 0.0058=5.43MeV. This amount of energy is primarily is used in
giving the kinetic energy of the alpha particle and energy of gamm photon which is emitted after
an alpha emission. When the ejected particles like alpha, beta or positron hits the surrounding
atmosphere, their kinetic energy is converted to heat energy. Like alpha particle, the emission of
beta or positron particle involves a mass defect.
β -emission:
14
14
C
N +
7
6
0
-1
e+
0
e
+ K.E of β
0
Δm = 0.0017 amu = 0.0017 X 931.6 = 1.58 MeV
In this case the decay energy is not sufficient to produce a gamma photon. So in many beta
emission processes is accompanied by the emission of another chargeless particle with negligible
mass called antineutrino(  ), which is truly anti electron neutrino. So decay energy is used
up in giving kinetic energy to the beta particle and forming anti electron neutrino.
Positron emssion:
11
C
6
11
0
B + +1 e +
5
0
e
0
In this case too, the decay energy is small like beta emission. Hence it is not sufficient for
producing a gamma photon. So many positron emission processes are accompanied by the emission
of another chargeless particle having negligible mass called neutrino(  ) which is truly electron
15
neutrino. So decay energy in this case is used up for giving kinetic energy to positron and
producing electron neutron. More on neutrinoes will be discussed later.
NUCLEAR FISSION:
Neutron induced nuclear fission is commonly understood as nuclear fission. When a heavier
radioactive nuclide like uranium-235 isotope(92U235) is bombarded upon by a slow neutron(0n1),
the heavier nuclide splits into a pair of lighter nuclides like 56Ba141 and 36Kr92 along with release
of 2 to 3 neutrons and a large quantity of energy. This is called nuclear fission and the energy that
we get is called the nuclear energy. This is a nuclear reaction but not a nuclear decay process.
In spontaneous decay, an alpha or beta or positron is emitted from the nucleus and the daughter
nuclide is also a heavy nuclide. It differs from the parent nuclide by maximum of two four
units(alpha emission) or no mass unit(beta or positron) and the energy released due to this is
small. But in nuclear fission, one subatomic particle, neutron is allowed to hit on a radioactive
nuclide like U235 by us. By that it splits the heavy nuclide into two halves forming much lighter
daughter nuclides having masses many units smaller than the parent nuclide.The energy released
in this process is enormously large. This nulcear energy can be constructively used to get electricity
in nuclear power plants and also can be destructively used to prepare nuclear bombs(atom
bombs). You know that in India there are many nuclear power plants in places like Kalpakam,
Tarapur etc. where electricity is obtained from nuclear power. The heat energy obtained during
the fission reaction boils water to produce steam and the steam rotates the turbine kept under a
magnetic field to produce electricity.
92
U235 + 0n1 ---> [92U236] --------> 56Ba141 + 36Kr92 + 3 0n1 + Energy
The slow neutron which is bomorbaded upon the U(235) nuclide has speed of 2.2 Km/sec.
having energy 0.025 eV. These are also called thermal neutrons because the speed of neutrons
at room temperature is nearly 2.2 Km/s. The heavier nuclide absorbs the neutron and is converted
to another isotope of uranium(U236). Since this isotope is highly unstable, it immediately splits into
two relatively lighter unclides having intermediate mass numbers(explained before) e.g Ba and
Kr alongwith 3 neutrons and a large quantity of fission energy. In reality several pairs of
daughter nuclides such as Cs/Rb, Xe/Sr, Te/Zr etc are formed alongwith Ba/Kr. But we
have given here only one pair (Ba/Kr) for simplicity in understanding.
235
92U
+ 0n1
[ 92U236 ]
90
38Sr
+ 54Xe143 + 3 0n1 +E
92
36Kr
+ 54Ba142 + 2 0n1 +E
52Te
137
144
55Cs
16
+ 40Zr97 + 2 0n1 +E
+ 37Rb90 + 2 0n1 +E
There altogether 300 isotopes(150 pairs of nuclides) from 37 elements which are formed from
the fission of U-235 ranging from Zn72 to Dy161. These nuclides are themselves radioactive and
are mostly beta emitters. They decay down to 80 stable isotopes belonging to 30 elements. For
example, Kr-92 decays by emitting successive beta particles to form stable istope of Zr-92.
36 Kr
92
-
37 R b
92
t 0.5 = 4.5 s
t 0.5 = 1.84 s
39 Y
92
-
t 0.5 = 3.54 h
-
40 Zr
38 Sr
92
-
t 0.5 = 2.71 h
92
(stable)
FISSIONABLE AND FISSILE NUCLIDES :
The above two terms are often used interchangeably for the same meaning. Actually they convey
different meanings. Nuclides which can be fissioned by neutron bombardment are called fissionable
nuclides. But out of those which can be fissioned by slow neutrons are called fissile nuclides.
Some examples of fissile nuclides are 92U235, 93Np239, 94Pu239, 94Pu241 and 92U233. These are
called nulcear fuels. 92U238 is fissionable nuclide but not fissile. It can be fissioned by fast
neutrons having speed of 52,000 Km/sec(14.1 MeV), not by slow/ thermal neutrons. Hence all
fissile nuclides are fissionable but the vice versa is not true. The other fissionable nuclide which
is not fissile is 90Th232. The above two fissionable nuclides namely U-238 and Th-232 are
17
converted to fissile Pu-239 and U-233 respectively by breeding process the details of which will
be discussed later. The use of fast neutrons to fission U-238 is not convenient as a chain reaction
is not possible for sustained use of the fuel. Hence for all practical purposes these two nuclides
are called non-fissile nuclides. But theoretically it is fissionable. Henceforth we shall not use
the term fissionable for U-238 and U-232 rather use the term non-fissile for them.
UNCONTROLLED AND CONTROLLED CHAIN REACTION:
Natural uranium minerals such as
pitchblend, monazite, clevite etc. contain
only 0.72% of fissile U-235 nuclide and
the rest non-fissile U-298(99.27%), of
course a very small quantity of nonfissile
U-234(0.005%). When the first slow
neutron bombards on the one U-235
nuclide, three neutrons are produced.
These neutrons produced are fast neutrons
having speed 20,000 Km/sec, but not
capable for further fission of both U-235
and U-238. These neutrons are slowed down by using a moderator like light water(H2O) or
heavy water(D2O) or graphite. If the % of fissile U-235 i shigh(enriched uranium) then the three
neutrons produced in the first fission step bombard on three more U-235 isotopes and each give
a pair of Ba and Kr and three additional neutrons + envery. So the bombardment of three
neutrons onto three U atoms generates 3 X3 =9 neutrons. These 9 neutrons bombard again on 9
fresh U-235 atoms to generate 9 pairs of daughter nuclides and 9X3=27 neutrons + heat. This
chain continues indefinitely. Thus the number of neutrons obtained increases in geometric proportion
and the fission reaction propagates in an uncontrolled manner. The time required for the production
of a netron in one step and its subsequent bombard on another fissile nuclide as small as 10–6 sec.
So within a very short time vast number of fissions take place to generate vast quantity of
energy. This is called uncontrolled branching chain reaction. The energy released during the
chain reaction is so vast that it is catastrophic and results in an terrible explosion(atom bomb)if
the reaction is not controlled. Control of chain reaction is done by using optimising the % fissile
nuclide in the nuclear fuel with respect to the moderator used. If natural uranium is used with
light water as moderator, then chain reaction cannot build up as most neutrons will be lost without
hitting a fissile nuclide. Some hit on abundant U-238 to convert to its heaver isotope which does
not fission. Some will escape out of the reactor. So after sometime the fission will stop.
U238 + 0n1 -------> 92U239
92
Controlled chain reaction is achieved by properly optimising the percentage of fissile material in
the fuel with respect to the moderator used. For light water moderator, enriched uranium containing
2.5 - 3.5% of U-235 is used. India imports enriched uranium ore to feed its light water based
nuclear reactors. Natural uranium is used as fuel when heavy water is used as moderator.
The optimum mass of fissile nuclide that should be present in the fuel to sustain the chain reaction
by allowing only one neutron per fission to undertake further fission and discard the other
neutrons is called critical mass of the fissle nuclide. In the side diagram, the fission giving two
18
neutrons per fission has been shown. One neutron is allowed for further fission each time and
other neutron is absorbed in the fuel by U-238 or otherwise. Slightly excess to critical mass of
fissile nuclide is often taken in the nuclear reactors. Cadmium and boron steel rods are also
used to absorb some excess neutrons and thus to control the fission reaction whenever required.
These are called controlled rods. When the number of neutron per fission undertaking further
fission goes slight above 1, then the control rod is lowered into fuel rods to absorb the excess
neutrons. When this number goes slightly below 1, then the control rods are raised away from
the fuel rods. To remove the great amount of heat energy produced during fission reaction,
coolants are used. Light and heavy water, liquid sodium, or sodium-potassium alloy etc. are
used as coolant to absorb the heat produced and remove the heat from the reaction site. Note
that the controlled chain reaction continues for years together(usually more than 10 years) in a
reactor to get sustained energy in nuclear power plants.
Critical, Supercritical and Subcritical Mass :
We have already discussed the importance of critical mass of the fissile nuclide for a controlled
chain reaction. If the percentage of fissile nuclide is higher than critical mass it is called
supercritical mass which goes for a uncontrolled chain reaction as the neutron per fission used
for futher fission reactions is above 1. High pecentage of U-235 or pure U-235 is used for the
purpose of making nuclear weapons(atom bombs). If the percentage of fissile nuclide is less
than the critical value, it is called subcritical mass which goes for stopping the chain reaction
after sometime.
Calculation of Fission Energy:
(a) Mass defect method:
The release of great amount of energy during fission reaction is due to the loss of mass(mass
defect) and its conversion into energy. Let us calculate this. The isotopic masses are written
below each species in the following equation.
U235 +
n1 ----------> 37 Rb 93 +
Cs 141 + 2 0n1
92
0
55
235.0012amu
92.9035amu
140.894
1
Mass of neutron(0n ) = 1.00885amu
Let us find the total mass of LHS which consists of one 92U235 atom and one neutron and total
mass of RHS which consists of one 537Rb93, one 55Cs141 and two neutrons.
Total mass of LHS :
235.0012 + 1.00885 = 236.01005 amu
Total mass of RHS:
92.9035 + 140.894 + 2X 1.00885 = 235.8152 amu
We find that the total mass of LHS is more than that of RHS. Thus some mass is lost during the
fission reaction. This mass loss has been converted to energy which is called the fission energy.
Mass loss = Δm = 2236.01005 - 235.8152 = 0.19485 amu
We know that loss of 1 amu mass gives an energy of 931.6 MeV,
Energy evolved by one 92U235 atom = 0.19485 X 931.6 = 181.52 MeV
Mass evolved by 1 mole i.e 6.023 X 1023 atoms of U = 181.52 X 6.023 X 1023 MeV
= 1.093 X 1026 MeV
Mass evolved by 1 gm of U atoms = (1/235) X 1.093 X 1026 MeV = 4.651 X 1023 MeV
= 4.651 X 1023 X 1.602 X 10–19 X 106 joules.
= 7.5 X 1010 J = 7.45 X 107 kJ
Thus we find that the fission of 1 gm of U(235) will release such a vast amount of energy. This
is incredible!!!
19
(b) Binding Energy Method:
The BE/N of the U-235 is 7.6 MeV while the average BE/N of the fragments having
intermediate mass numbers(40-140) is 8.4 MeV. This difference of binding energy is released as
fission energy during fission reaction.
Total binding energy of one U-235 nuclide = 235 X 7.6 = 1786 MeV
Taking 37Rb93 and 55Cs141 in the reaction shown above, and using the average BE/N
of 8.4 MeV, the total binding, the total binding energy of the fragment pair = 8.4 X ( 93+141)=1965.6
MeV
Difference in Binding Energy(Fission energy) per nuclide = 1965.6 – 1786 = 179.6 MeV
This value closely matches with the that calculated by using the mass defect method above. This
small difference in the results obtained by the above two methods is that in the second method,
the average BE/N was used not the exact ones for the pair.
Note that the two methods for calculating the fission energy are in principle the same because
the mass defect is responsible for the binding energy difference.
SAQ 12:
What other pairs of daughter nuclides are obtained when U235 is bombarded
upon by neutron during fission reaction.
SAQ 13:
Between the spontaneous alpha emission of U235 and fission reaction by neutron
bombardment which process releases greater energy and why?
PRESSURIZED LIGHT WATER REACTOR(PLWR):
There are many type of nuclear reactors out of which the common reactors in India are
(a) Pressurized Water Reactors(PWR)
(b) Boiled Water Reactors (BWR)
(c)WWER/ VVER (Water Cooled Water Moderated Energy Reactor) VVER
is the Russian abbreviation.
In India PWR and VVER are in use. There are two types of PWRs namely (i) Pressurized
Lighte Water Reactor(PLWR) and (ii) Pressurized Heavy Water Reactor(PHWR). In India all
all the nuclear reactors are of PHWR type and the plant at Kudankulam is of VVER type.To
understand the functioning of nuclear reactor easily we shall take up PLWR first.
20
PLWR
PHWR
PLWR : Light water(H2O) is used as coolant-cum-moderator in this reactor which uses enriched
uranium ore(2.5-3.5% U-235). It consists of fuel rods installed vertically in the core of a thick
cylindrical steel vessel suspended in light water present in the vessel. Each fuel rod is 4 m in
length and 1 cm in diameter made of steel or zirconium alloy. This contains the UO2
pellets(enriched), the nuclear fuel. About 200-300 fuel rods are kept together to form one bundle.
About 150-200 such bundles are kept in a vertical position inside the cylindrical vessel(1 m
diameter adn 1.7 m long) into which the control rods are suspended from upwards. Water acts
both as moderator(slowing down the speed of neutron) and also coolant. Since water is kept
inside the vessel at 150 atmosphere at 3250C, it is called pressurized water reactor. Water does
not vaporize at this perssure and has a much enhanced capacity to carry heat. There are water
inlet and exit tubes connected to the vessel. The water carrying the heat produced in the reactor
21
comes out through the exit tube (RHS) and returns back to the to the reactor through the inlet
tube after giving away the heat in the heat exchanger. This cyclic rotation of water takes place
continuously in the reactor. Water in the exchanger gets converted to steam which runs a turbine
which rotates a coil under magnetic field. This electricity is generated. Controlled rods(cadmium
rods or boron steel rods) are suspended from above which is made up and down as and when
required. When the heat produced becomes more than normal(heading towards uncontrolled
chain reaction when neutron per fission goes above 1), the control rods are lowered into the fuel
rods so as to reduce the neturons and control the chain reaction. When the heat produced
becomes less than the normal(when the number of neutron per fission goes below 1) the control
rods are raised up so that more neutrons are allowed to undertake fission and sustain the chain
reaction.
PHWR: Heavey water(D2O) is used as moderator-cum-coolant in this reactor with unenriched
i.e natural uranium as fuel. The basic principles of functioning of PHWR is same as PLWR.
Except that the technology is such the reactor can be refuelled without shutting down the reactor.
There are a large number of pressure tubes kept in horizontal positions in which fuel is kept as
well as heavy water is cycled. We shall not go into the details of PHWR here. In India most of
the reactors are PHWR. There are two BWR plants in Tarapur, Rajasthan.
Starting of Fission Process : A cource of neutron is brought near the fuel rods in the reactor.
The source of neutron consists of a mixture of beryllium powder and an alpha emitter like
Ra226 or 98Cf252. Alpha particles bomard on a Beryllium nuclide to form a neutron and a carbon
88
nuclide as shown below. This is called nuclear transmutation about which we shall study later.
He4 + 4Be9
→
n1 + 6C12
0
Once the fission process starts, the neutron source is removed from the reactor as the reaction
becomes self sustaining.
2
FAST BREEDER REACTORS (FBR)
In fast breeder reactors, we kill two birds in one stone. We get nuclear energy like PWR/VVER/
BWR, at the same time we get more quantity of fresh fissile nuclide for future use. This is
achieved by using highly enriched Uranium(15-30% U-235) and fast neutron to bombard on it.
Since breeding of new fissile materials is done here and fast neutrons are used for the purpose,
it is called fast breeder reactors. The difference between the conventional nuclear reactors
explained before and fast breeder reactors is as follows.
(a) FBR uses fast neutrons while conventional reactors use slow neutrons
(b) FBR does not use any moderator like LW or HW or graphite while others do.
(c) Liquid sodium or Na-K alloy is used as coolant in FBR while in others water is often
used as coolant.
(d) FBR uses highly enriched(15-30% U-235) uranium while others use either natural or
slightly enriched(2.5-3.5%) uranium.
FBR consists of two parts in the core of the reactor.
(i) Interior part : This part consists of nuclear fuel rods i.e enriched
uranium. Fission starts here to generate fresh fast neutrons(at an average of 2.4 per fission)
having 20,000 Km/s speed. Out of them, one is used to undertake further fission to sustain the
controlled chain reaction and generate sustained energy and the rest 1.4(average) escape out of
22
the interior part to hit on the exterior part.Note that at such high percentage of fissile nuclide, fast
neutrons can bring about controlled chain reaction. Slow neutrons are not required for the purpose
and hence moderator is not used. Like a conventional reactor, the energy formed in this part
produces steam which runs a turbine to generate electricity.
(ii) Exterior part : This part surrounds the interior part as a blanket.
This part contains natural uranium(0.7% U-235) or depleted uranium( < 0.7% U-235) in the
form of rods. The excess 1.4 neutron per fission which escape out of the interior part hit on U238 present in this part to convert finally to 94Pu239 with two successive beta emissions as given
below.
92U
238
0n
1
fa st n e u tro n
239
92U
t 0 .5 = 2 3 .5 m in
-
-1 e
0
93N p
239
t 0 .5 = 2 .3 5 d a ys
-
-1e
0
94P u
239
First U-238 changes to its isotope U-239 by absorbing the fast neutron. U-239 is a beta emitter
which changes 93Np239 after emitting a beta particle. The latter is also a beta emitter which
changes to 93Np239. The latter two nuclides have small half life periods so as to quickly form the
fissile 94Pu239 which has a long half life of 2.44 X 104 years. This fissile 94Pu239 nuclide is stored
for a later use in conventional power plants. It has been estimated that for every 100 fissile U235 used up in the interior part for getting nuclear energy 115 fissile Pu-239 are produced in the
exterior part for future use. Is it not very interesting !!!!
Just like breeding of fissile 94Pu239 is done from U-238, breeding of another fissile nuclide U233 is done from 90Th232 as per the following.
232
90Th
1
0n
233
90Th
-
0
-1e
233
91Pa
-
0
-1e
233
92U
fast neutron
In India, commissioning of 500 MW PFBR(Prototype FBR) at Kalpakam will be ready by 2013.
8 more FBRs are on the pipeline at the same place. A 13 MW test FBR is successfully working
there since 1985. The 500 MW FBR will make use of blend of Pu-239 and U-235 (their oxide)
in uranium spent fuel from conventional plants as fuel. So Pu-239 will be burned in the interior
part and depleted uranium will be used as usual in the exterior part to get more Pu-239 in this
case. So less Pu-239 is burned and more Pu-239 is breeded. Since the fission of U-233 in
nuclear plants is less hazardous,breeding of U-233 from Th-232 is considered as the future of
nuclear power in the FBRs.India is the 6th nation to have FBR technology.
SOME FREQUENTLY ASKED QUESTIONS(FAQs) ON FISSIONAND THEIR
ANSWERS:
FAQ 1 : What is spontaneous nuclear fission and how it is different from neutron induced
fission.
Answer : There are two types of nuclear fission. One is neutron induced which has
been discussed in details. The other is spontaneous. Some heavier unstable nuclides having mass
numbers much greater than 230 spontaneously fission into two fragments, some neutrons and
energy.
Cf252
92
→
56
Ba142 + 42Mo106 + 4 0n1 + energy
FAQ 2 : Why it is not convenient to get nuclear energy from U-238 by using fast
23
neutrons.
Answer : Fast neutron necessary for fissioning U-238 should have a speed of 52,000
Km/sec. But the neutrons which are obtained from fission, although fast(20,000 Km/s) but not
able to undertake further fission. Hence a self sustaining chain reaction is not possible in this
case. Hence it is not used in generating nuclear power.
FAQ 3 : Why all the fissile nuclides have odd mass numbers ?
Answer : Have you marked that all the fissile nuclides carry odd mass numbers like U235, U-233, Pu-239. But nuclides having even mass numbers like U-238 is non-fissile for all
practical purposes. The reason is - in nuclide having odd mass number one neutron remains
unpaired. When it absorbs another slow neutron, the already existing unpaired electron pairs
with the incoming electron and thereby releases some quantity of energy. This is called pairing
energy which is 6.8 MeV. Fission of the nucleus needs 6.5 MeV (activation energy) which is
less than the pairing energy. Hence fission becomes possible and after fission, we get much
more quantity of energy as discussed before. In U-238, there is no unpaired electron. Hence
there is no pairing energy in this case. The energy released when it absrobs a neutron is 5.5
MeV which is insufficient to provide the activation energy for fission of U-238 i.e 7.0 MeV.
Hence fission is not possible by slow neutrons. But by using very fast neutrons possessing very
high K.E, it is possible to overcome the activation energy of U-238 fission. But this is not
commercially viable for the reason already discussed.
FAQ 4 : When boron steel rods(control rods) are hit by excess neturons what change
occurs ?
Answer : The following nuclear transmutation occurs when neutron bombards on boron
steel. We get an alpha particle alongwith Li-7 isotope.
5
B10 + 0n1
→
3
Li7 + 2He4
FAQ 5 : Where are the heavy water plants present in India ?
Answer : There are 7 heavy water plants in India - Nangal, Boroda, Tutikorin, Kota,
Thal, Hazira and Manuguru.
NUCLEAR FUSION
This is just opposite process of nuclear fission. In nuclear fission that we discussed before, a
very heavy nuclide like U235 is broken down into two relatively lighter nuclides(having intermediate
mass numbers) and energy is obtained during the process. In case of nuclear fusion, two very
light nuclides (eg. 1H1, 1H2 , 1H3) combine or fuse to form relatively heavier nuclides(eg. 2He3,
He4) thereby releasing very great amount of energy. This is called fusion reaction and the
2
fusion energy released is much greater than the fission reaction of U235 isotope.
1
H1 + 1H3 → 2He4 + energy
Refer the BE/N – mass number graph given before few pages. The graph rises sharply but falls
gradually. Hence binding energy difference between very light nuclides such as protium, deuterium
or tritium and the product nuclides like He-3 or He-4 is much greater than that we found for
fission reaction. Like fission reaction, the binding energy difference is released, in this case, as
fusion energy. We can also explain this on the basis of mass defect. Like the fission reaction, in
fusion reaction too, there is a large loss of mass for which the energy released is enormously
large. The total mass of the reactants(LHS) is greater than the total mass of the products(RHS)
24
and the difference in mass is lost in the form of fusion energy.
Disadvantages in Fusion Reaction :
(1)
Look to the difference between fission and fusion. In fission, a neutral particle(neutron)
was bombarded upon a heavy nuclide like U-235. So there is no difficulty in its absorption by the
nucleus. But in fussion, two positively charged nuclei will have to fuse with each other. For that
they have to overcome huge repulsive force between them. Therefore this becomes possible by
applying very high temperature of the order 107 – 108 0C as well as very high pressure. At high
temperature, the kinetic energy of particles would be enormously large and if high pressure is
applied simultaneously they would come very close to each other to undergo fusion. So far as
temperature is concerned, high temperature is applied initially by external method, but once
fusion reaction gets started, the heat produced during the reaction becomes self sustaining and
temperature is maintained due to that. No more heat is required from external source.
(2)
The principal nuclide for fusion is hydrogen. Since it is a combustible gas, there is always
a great amount risk involved in using hydrogen. That is why, fusion reactions have not yet been
tamed to the extent of using it in a commercial basis. Only it has been a great subject matter of
research. But the efforts of the scientists are on to tame down the reaction and once it is
achieved, the energy crisis will be all over.
Fusion in Sun and other Stars:
The energy produced in sun and stars is due to continuous fusion reactions taking place there.
[ 1H1 + 1H1 ---------> 1H2 + +1e0 + 1.44 MeV ] X 2
(step -I)
2
1
3
[1H + 1H ----------> 2H + 5.94 MeV ] X 2
(step-II)
3
3
4
1
H
+
H
-------->
H
+
2
H
+
12.86
MeV
(step-III)
2
2
2
1
__________________________________________
4 1H1 ----------------> 2He4 + 2 +1e0 + 26.72 MeV
In the first step two hydrogen (protium) nuclides(protons) fuse to give a deuterium nuclide, a
positron and fusion energy of 1.44 MeV. In the second step, the deuterium nuclide formed in the
first step fuses with another protium nuclide to form a helium-3 nuclide(2He3) and again fusion
energy of 5.94 MeV is released. In the third step, two 2He3 nuclides(formed in the second
step)fuse to give one 2He4 nuclide and two protium(1H1) nuclides and again fusion energy of
12.86 MeV. The sum of these three steps can be obtained by cancelling the 1H2 and 2He3 after
making necessary multiplications(2 multiplied with each of first and second steps). Thus we find
that a great amount of energy(26.72 MeV) is released in all the three fusion reactions taking
place in sun. The energy released during the formation of one mole of He-4 from hydrogen in
fusion reaction ( 26.72 X 6.023 X 1023 = 160.9 X 1023 MeV) is 9 times greater than the fission
energy obtained from 4 gm of U-235. It has been estimated that sun has 1033 gm of hydrogen
gas. Only 2% is exhausted in every 109 years.Hence to exhaust all the hydrogen present in the
sun, time required will be 1010 years.
Advantages of Fusion Reactions :
(1)
Nuclear fuel for this reaction is hydrogen, which can be obtained in inexhaustible quantity
from sea water in a cheap price compared to Uranium-235 used for fission.
25
(2)
The energy obtained from this much greater.
(3)
It is more safe as very less hazardous radioactive emissions take place from the fusion
reactors.
Some other Fusion reactions:
1
→ 2He3 + 0n1 + Energy
H2 + 1H3 → 2He4 + 0n1 + Energy
1
H2 + 1H2
The fusion of H-2(deuterium) with H-3(tritium) has been possible but deuterim-deuterium
fusion has not been possible yet. Because it requires much higher temperature. If this fusion
could be possible, then it will be more convenient for getting energy. Deuterium is obtained
from ordinary source of hydrogen such as water. But tritium is synthetically prepared with
much difficulty as it is not available in nature.
FUSION REACTORS:
High temperature in fusion reactors is achieved by
(a) microwave furnce
(b) laser beam
(c) Fas neutron bombardment
In Fast Neutron Bombardment(FNB) process, fast neturons are bombarded on hydrogen gas
which not only bread the bond and make them atoms but also strip off the electron to convert
them to hydrogen ions or protons. This high velocity collision of neutron beam with hydrogen
produces also a very hight temperature required for fusion. At such high temperature, all the
species exsit as ions. This is called the plasma state of matter.Superconducting magnets are
used to compress or apply very high pressure onto the plasma to bring about fusion.
Hydrogen bomb:
Hydrogen bomb makes use of fusion reaction. Hydrogen bomb is much more devastating than
atom bomb which makes use of fission. The fusion reaction involved in the hydrogen bomb is
given below.
H2 + 1H3 ---------> 2He4 + 0n1 + 17.6 MeV
1
Here one deuterium(1H2) nuclide fuses with one tritium(1H3)nuclide to form one helium(2He4)
nuclide and a lot of energy. Hydrogen bomb is also called a thermonuclear bomb. The high
temperature needed for the fusion(107 - 108 0C) is obtained by an intial fission reaction. Once
fusion starts, it becomes self sustaining.
In fact tritium is obtained synthetically by the neutron bombardment on Li-6 nuclide. Therefore
for hydrogen bomb, Lithium(6) Deuteride(Li6D) is used to which neutron is bombarded.
6
+
1
0n
3
1H
+
3
1H
+
2
1H
2He
4
3Li
3Li
6
+
2
1H
+ 2He
+
4
1
+ 0n
+ Energy
+ Energy
2 2He4 + 22.4 MeV Energy
Tritium obtained in the first step fuses with deterium to form He-4 and lot of energy.
The first hydrogen bomb test by USA in 1952 in Eniwetok island which was based on the above
26
reaction. Now several countries including India has the know-how of a hydrogen bomb.
Artificial Nuclear Transformation or Transmutation and Artificial
radioactivity:
When highing accelerated subatomic particles like neutron, proton, alpha particles etc.are allowed
to bombard onto a nuclide of an element, it is transformed into a nuclide of another element.
Thus a new element is produced. This is called nuclear transformation or transmutation. Most of
the synthetic elements(transuranic elements, tritum, Tc, Pm, At, Fr, Co-60, C-11, Br-82, Na-24
etc) have been prepared by this technique. The first nuclear transmutation was studied by
Chadwick who discovered neutron. When beryllium metal was bombarded upon by alpha particles
coming from a radioactive alpha emitter, it was surprisingly found that silvery white beryllium
metal was slowly transformed to a black nonmetallic substance, carbon. In this process a beam
of neutral particles was emitted. These particles were named as neutrons.
Be9 + 2He4 ---------> 0n1 + 6C12
4
Add the mass numbers shown in superscripts in LHS species and RHS species separately you
will find them equal. Also add the atomic numbers in subscripts in LHS and RHS which would
also be equal. In fact we can know the element formed during the transformation reaction if we
know the element onto which bombarded, particle which bombards on the element and the new
particle produced in the process. Look to the following process.
Let us bombard neutron(0n1) onto 8O16 nuclide. We find that a new particle is produced.
This new particle is an alpha(2He4) particle. What is the new element formed in the transformation
process?
O16 + 0n1 ---------> ? + 2He4
8
Let us add the atomic numbers in the LHS species given in the subscripts. It is 8 + 0 =8. In the
RHS, we know the atomic number of alpha particle is 2. So the remaining atomic number in
RHS would be 8 – 2=6. Now what is the element having atomic number 6? It is carbon. So
carbon is the new element formed. The mass numbers are then added in LHS. It 16 + 1 = 17. In
the RHS, an alpha particle has a mass number of 4, so the remaining mass 17 – 4 =13, will be
mass number of the carbon formed. So it is 6C13 isotope.
O16 + 0n1 ---------> 6C13 + 2He4
8
Thus we find that ordinary gaseous oxygen when bombarded by neutrons is converted to solid
carbon(not the usual C12 isotope but the uncommon C13 isotope). This is interesting indeed, just
like a magic !!
The above reaction shown(oxygen to carbon conversion) is called a 8O16(n,α) reaction. The first
particle given inside the bracket(n) is the particle bombarded and the second particle(α) is the
particle formed during the transformation process. The target element undergoing transformation
is written first. This type of convention is used in all nuclear transformation processes.
Transformation by alpha particle:
F19 (α,p) reaction :
F19 + 2He4 --------> 1H1 + 10Ne22
9
9
33
As75 (α,n) :
Transformation by protons:
P31(p,n) reaction :
15
27
→ 35Br78 + 0n1
33
As75 + 2He4
15
P31 + 1H1 ----------> 0n1 + 16S31
13
3
Al27 (p,γ) :
Li7 (p,α) :
→ 14Si28 + 0γ0
Li7 + 1H1 → 2He4 + 2He4
3
13
Al27 + 1H1
Transformation by neutrons:
13
3
Al27 (n,p) :
Li6 (n,T) :
O16 (n,α)
8
Transformation by deuteron:
Li6 (d,p) reaction:
3
13
Al27 (d,α) :
→ 12Mg27 + 1H1
Li6 + 0n1 → 1H3 + 2He4
3
O16 + 0n1 → 6C13 + 2He4 ;
8
13
Al27 + 0n1
Deuterium ion (1H2) is called a deuteron.
Li6 + 1H2 ---------> 1H1 + 3Li7
3
13
Al27 + 1H2
→ 12Mg25 + 2He4 ;
Transformation by gamma rays:
Be9 (γ,n) :
Be9 + 0γ0 → 2 2He4 + 0n1
4
Transformation effected by heavier projectiles: For the preparation of many synthetic isotopes
4
most of which are radioactive, heavier nuclides like Li-6, Be-9, B-11, C-12, O-16, F-19, Ne-20
etc. have been used as projectiles to hit still heavier particles. By this method, transuranic elements
also have been synthesized.
→ 86Rn205 + 6 0n1
Au197 + 7N14 (projectile) → 35Br74 + 3 0n1
79
U238 + 8O16 (projectile) → 100Fm250 + 4 0n1
92
29
Cu65 + 6C12 (projectile)
SAQ 14:
Write the following nuclear equations as indicated.
(i) (α,p) reaction of 7N14
(ii) (α,n) reaction of 13Al27
7
(iii) (p,n) reaction of 3Li
(iv) (d, n) reaction of 83Bi209
SAQ 15:
Fill in the blanks:
50
(i) 24Cr + 2He4 --------> ? + 0n1
(ii) 42Mo96 + ? ---------> 43Tc97 + 0n1
7
1
4
(iii) 3Li + 0n -------> 2 2He + ?
(iv) ? + 2He4 --------> 12Mg26 + 1H1
14
1
14
(v) 7N + 0n ------->6C + ?
ARTIFICIAL RADIOACTIVITY:
Artificial radioactivity is an usual consequence of nuclear transformation that we studied before.
Look to the following (n,p) reaction of 13Al27 giving 12Mg27.
Al27 + 0n1 ---------> 1H1 + 12Mg27* (radioactive)
13
You know that the stable isotope of Mg is 12Mg24. So 12Mg27 contains excess neutrons than
expected(n/p ratio is more). So this nuclide is not stable and is radioactive. Can you guess what
particle it emits? Yes, it is a beta emitter. You know that the nuclides which have greater n/p
ratio(which lie above the band of stability) are beta emitters. The beta disintegration reaction of
Mg27 is as follows.
12
Mg27 ----------> -1e0 + 13Al27
(Artificial Radioactivity)
12
It decays with a fixed half life period to stable 13Al27 nuclide. This is called artificial radioactivity.
Let us take another example of (α,n) reaction of 13Al27 nuclide.
Al27 + 2He4 -------> 0n1 + 15P30* (radioactive)
13
28
The 15P30 nuclide produced is not a stable nuclide as P31 isotope is stable. 5P30 has less number
of neutrons than required. Since its n/p ratio is less than expected, could you guess what particle
it would emit in its radioactivity? Yes, it would emit a positron. The nuclides having lower n/p
ratio than expected are positron emitters.So the 15P30 isotope produced is radioactive and decays
as follows.
P30 -------> +1e0 + 14Si30(stable)
(Artificial Radioactivity)
15
SAQ 16: Give the (n,p) reaction of 7N14 and comment on the artificial radioactivity of the
product. Give the decay reaction of the product.
RADIOACTIVE SERIES:
All the natural radioactive elements and their isotopes belong to four series or families. Each
series is named after the first isotope from which the serites originates. For example, in case of
uranium series the first istope is U-238 which is an alpha emitter. It forms the daughter element
Th-234 after emitting an alpha particle. Th-234 is a beta emitter which forms the daughter
nuclide Pa-234. Pa-234 is also a beta emitter which forms another daughter nuclide. This process
continues till the last nuclide obtained is a stable one. In case of uranium series it is Pb-206. The
half life periods of each radioactive nuclide in the series are different. For exampel U-238 has
t0.5 = 4.5 X 109 years while Th-234 has t0.5 = 24.21 days.
(1)
Uranium Series / (4n + 2) Series :
92U
238
- 
90T h
234
- 
91P a
234
82P b
206
The first nuclide is U-238 and the last stable nuclide is Pb-82. This series contains 14 radioactive
nuclides. 8 alpha and 6 beta particles are emitted throughout the series. Since the mass numbers
of all the nuclides in this series is expressed as equal to (4n+2), where n is a whole number, it is
called (4n+2) series. For example: 238 = 4n + 2, so n = 59(whole number), 206 = 4n + 2, so n =
51.
29
(2)
Thorium Series / 4n Series:
90T h
232
-
88R a
228
-
89A c
228
82P b
208
In this series, the starting nuclide is Th-232 and last stable nuclide is Pb-208. This series contains
about 11 radioactive nuclides. At some places branching reactions are found. For example, Bi212 decays both by alpha and beta emission to give two different daughter nuclides which
ultimately give the same stable Pb-208 nuclide by emitting beta and alpha particles respectively.
Six alpha and 4 beta particles(excluding the branching reaction) are emitted during the series. All
30
the mass number of this series are expressed as 4n where n is a whole number. 232= 4n, so n=58
(3)
Actinium Series/ (4n+3) Series:
89Ac
227
82Pb
207
This series begins with Ac-227 nuclide and ends in the stable Pb-207 nuclide. The series has
some nuclides above Ac-227. It actually begins from U-235 which is an alpha emiiter and gives
Th-231. Th-231 is a beta emitter and produces Pa-231. Pa-231 is an alpha emitter and produces
Ac-227 which regarded as the starting nuclide for all practical purposes in this series. Actinium
series is also called U-235 series as the true starter is U-235. Since all these nuclides above Ac227 are scarce, it is popularly called actinium series. There are branching reactions at many
places in this series. Excluding the braching reactions, 5 alpha and 3 beta particles are emitted in
the series. All the mass numbers of this series are express by (4n+3) where n is a whole numberl
227 = 4n +3, So n = 56.
4.
Neptunium Series / (4n+1) Series:
93 Np
237
83 Bi
209
In this series, the starting nuclide Neptunium-237 is almost extinct in nature. But all other members
are available. The ending stable nuclide is Bi-209.
31
There is branching reaction at one place in this series. Excluding this, the number of alpha and
beta particles emitted during the series is 7 and 7 respectively. All the mass numbers are expressed
as (4n+1). 237 = 4n + 1, So n = 59.
SAQ 16 : Prove that in Uranium series 8 alpha and 6 beta particles are emitted.
Application of radioactivity:
The discovery of nucleus from the famous gold foil experiment by Rutherford was possible by
the alpha particles emitted from radioactive element. Neutron was discovered by Chadwick by
the help of radioactivity. Isotopes and isobars of elements were discovered first with radioactive
elements. The artificial elements beyond Uranium(transuranic elements) are the result of
radioactivity. The nuclear energy obtained in nuclear power plants is used to get electricity. By
carbon dating we could know the age of the old carbon containing material and hence the age of
many perished civilisations. Gamma rays are used to cure cancer. Other rays are used to detect
many other diseases. The leakage in the underground oil pipe lines, water pipes, gas pipe lines
can be detected through a technique called tracer analysis. A radioactive element is introduced
in the pipe line and its appearance is detected at suspected places through instruments like
Geiger Muller Counter(which detects and counts radioactive emissions). This helps to know the
point of leakage. The tracer analysis is also used in hundreds of chemical reactions to study the
pathway through which a reaction proceeds. Without dealing with the details, we list out here the
areas where radioactivity is used.
Diagnosis and Treatment of Diseases:
(1)
32
Several radioisotopes are used for diagnosis and treatement of many diseases. I-131 isotope is
used for diagnosis and treatment of diseases of thyroid gland. Co-60 is used for cancer, P-32 for
leukemia(blood cancer) and brain tumour, Na-24 for heart diseases. Technetium(Tc)-99 is used
for taking pictures of heart, lungs, liver, spleen etc. and diagnose diseases. Yttrium(Y)-90 is used
for liver cancer while Cu-64 is used to diagnose genetic diseases.
(2)
Food Industry :
Food can be preserved for a long time by using radioactive emission which kill microorganism
such as virus, bacteria etc.
(3)
Preservation of Antique Monuments:
By applying the radioactive rays, the antique monuments are being preserved from insects and
other microorganism.
(4)
Industry : Fabric, Paper, rubber, paints industries make use of some radioactive isotopes
to determine the quality of the products, thus used in quality control measures.
(5)
Agriculture :
Radioisopes are used with the fertilisers to carry out research on nutrition of plants. The details
we shall not discuss here.
(6)
Age of old organic matter (Carbon dating) :
The age of an old civilization could be estimated by carbon dating which has been explained
before.
(7)
Age of Earth:
From the ratio of U-238 and Pb-206 present in the uranium minerals, the age of the earth is
determined. This called Uranium-lead dating. We shall not discuss more this.
(8)
Tracer Analysis:
already discussed.
(9)
Sterelization of medical instruments:
The equipments and instruments used in hospitals can be sterelized by using gamma radiations
obtained from a radioisotope.
(10)
It helps in maintaining the temperature below earth crust. The radioactive emissions
produce a lot of heat when arrested inside earth crust.
(11)
Nuclear Fission and Fusion:
Already discussed before.
Hazards of Radioactivity:
Radioactive isotopes and the emitted rays exist in the earth crust as well as atmosphere. The
radioactive rays; alpha, beta, gamma are dangerous to human beings. Our food also contain
traces of radioactive substances. When these remain below the human tolerance limit, it is not
harmful. If it goes above that level can cause dreadful diseases like cancer in blood, skin, breast,
liver, colon, bladder, stomach, intestine etc. The radioactive Rn is present much above the limit in
the atmosphere of USA and other developed nations which has caused a great concern for
them. Human bodies contain radioactive K-40, C-14, Pb-210. Tobacco cotains Pb-210. The
flourescent tubes of television contains tritium.A lot of radiactive isotopes and their emissions
mix up the atmosphere from the nuclear waste in nuclear power plants, atom bomb test explosions.
This has caused a great threat to human civilization and WHO is deeply worried about this
33
nuclear pollution and engaged in combatting it.The persons working in the nuclear plants are
more exposed to the harmful effects of the radiations. Precautions should be taken to protect our
living environment from the harmful effects of radioactive rays.
Rudimentary Idea on Elementary Particles and Antiparticles:
The entire universe is made up of two types of elementary particles. Note that the proton and
neutron are also further divisible to still smaller particles. In other words they are made up of
elementary particles which are still smaller than them. These two types of elementary particles
are
(a) QUARKS
(b) LEPTONS
QUARKS :
There are six types of quarks.
(1) UP ( U quark)
(2) DOWN (D quark)
(3) CHARM (C quark)
(4) STRANGE (S quark)
(5) TOP (T quark)
(6) BOTTOM (B quark)
U, C and T quarks carry +2/3 units of charge while D, S adn B quarks carry –1/3 unit of
charge.Quarks always remain in groups. Hence they are social beings.
LEPTONS:
There are 6 types of leptons.
(1) ELECTRON (e)
(2) MUON (μ)
(3) TAU (τ)
(4) ELECTRON NEUTRINO (νe)
(5) MUON NEUTRINO(νμ )
(6) TAU NEUTRINO (ντ)
Each of electron, muon and tau has –1 of charge and the corresponding neutrinos have no
charge. So the electron about which you know so much is a lepton. Leptons always remain
isolated(not in group), hence they are not social beings.
(The beginners are advised not to confuse neutron with neutrino. The are different. While
neutron is a baryon made up of quarks, neutrinos are leptons)
ANTIPARTICLES:
Every particle has its antiparticle. Particle and antiparticle are identical in all properties except
one property. For charged particles, particle and antiparticles have opposite charges. Excepting
three neutrinos, the antiparticles of all other elementary particles have opposite charges w.r.t
their corresponding particles. For example, U quark has +2/3 charge, while anti U quark( U )
has
–2/3 charge and so on. The antiparticles are denoted with bar sign above the symbol
of their particles. For neutrinos, the antiparticles too have no charge. But they differ in more
34
than one properties, principal among them is difference in masses. We shall not discuss more
about this here. Just remember that antiparticle is like the reflection of reflection of a particle
which are identical in all respects except charge(valid for charged particles).
Note that there are six antiquarks.
(1) U (Anti U)
(2) D (Anti D)
(3) C (Anti C)
(4) S (Anti S)
(5) T (Anti T)
(6) B (Anti B)
There are also six antileptons.
(1) e (Anti e, commonly called a positron with symbol +1e0)
(2)  (Anti μ, this has also +1 charge but has a mass)
(3)  (Anti τ, this has also +1 charge but has a mass)
(4)  e (Anti electron neutrino, commonly called neutrino. This has no
charge but negligible mass like electron)
(5)   (Anti muon neutrino. It has no charge but has mass)
(6)  (Anti tau neutrio. it has also no charge but has mass)
So altogether there 12 particles (6 quarks and 6 leptons) and also 12 antiparticles(6 anti quarks
and 6 antileptons). The entire universe is composed of these 12 particle/antiparticle pairs.
Antiparticles are unstable and hence do not exist freely. The common antiparticle which is
emitted from some radioactive nuclides is positron i.e antielectron. The presence of all other
antiparticles have been proved beyond any doubt now.
ANNIHILATION:
The interesting fact about particle-antiparticle is that when any particle comes in contact with its
own antiparticle, both the two are destroyed and their mass is converted to energy. It has been
already told that when a positron(antielectron) comes in contact with an electron, both electron
and positron destry each other and we get two gamma photons moving in opposite directions.
This is called annhilation process. The same is the case with any particle annihilating with its
own antiparticle. Just like a particle has its antiparticle, a bigger particle like proton, helium atom
etc.i.e a matter has its antimatter.
Recently scientists have prepared antiproton, antihydrogen and antihelium atoms. They
are extremely unstable having very small life periods. Scientists believe that every matter can
have its antimatter, though not available, can be prepared.
It is believed that when the universe was formed by the big bang, equal number of matter and
antimatter were formed. Immediately after these matter and antimatter underwent annhilation
to pure energy. But some matter was left out. In other words there was an imbalance between
35
matter and antimatter after big bang. The universe we see is that left-out matter. So scientists
believe that there must be the corresponding antimatter somewhere hidden which we do not
find. Often you might have heard the fantasies/humour about this antimatter. Some are excited
to believe that there must somewhere an ANTIWORLD where every matter here has an anti
version. For example, Anti Sonia must be reading in an anti high school and eating anti ice cream
and riding anti cycle and so on. If Sonia living in this world somehow comes in contact with Anti
Sonia living in antiworld, can you imagine what will happen ? Both Sonia and antisonia will
destroy each other and covert to pure energy !!! This is merely fantasy.
HADRONS:
Quarks always remain in groups, not isolated. They remain either as a combination of 2 quarks
or 3 quarks. These combinations of quarks are called HADRONs. Hadrons are of two types.
(1) BARYONS : combination of 3 quarks
(2) MESONS : combination of 2 quarks.
Examples of Baryons : Proton and Neutron which you know are baryons
PROTON : UUD (it is combination of two U quarks and one D quarks.
So count the total charge. Is it not +1 ? +2/3 + 2/3 – 1/3 = +1
NEUTRON : UDD (its a combination of one U quark and two D
quarks. Is it not netural ? +2/3 – 1/3 – 1/3 = 0
Examples of Mesons: There are many types of mesons out of which the π mesons
also called pions are more common. There are three types of pions.
(1) π+ (positve pion) : U D (it is combination of one U quark and and anti D
quark.
(2) π– (negative pion) : D U ( it is a combination of one D quark and an anti
U quark). Note that π+ is the antiparticle of π–.
(3) π0 (neutral pion) : U U or D D (it is made up of U or D quark with
its own antiparticle. Note that there is no annhiliation occuring there as long as a neutral pion is
in life.
FUNCTION OF PIONS:
(1)
STABILITY OF NUCLEUS(MESON THEORY):
The most important function of pions(pi mesons) is to bind the nucleons(neutrons and
protons) to form a stable nuclues.
When a neutron emits a π– , it is converted to a proton and when another proton absorbs
a π– , it is converted to a neutron. Similarly when proton emits a π+, it is converted to a neutron
and when another another neutron absorbs a π+ , it is converted to a proton. Hence the charged
pi-mesons rapidly exchange beween proton to neutron and neutron to proton. Neutral pions are
exchange between proton to neutron and vice versa. When a proton emits a π0, it remains as a
proton and when a neutron absorbs π0 it remains as a nuetron. Theses rapid exhange of pions
between the nucleons is responsible for the binding of nucleus and hence its stabililty.
In fact, when a U quark changes into a D quark in a proton, it is converted to neutron
during beta emission of a radioactive nuclide, (UUD changes to UDD), another particle called
W– boson is emitted first, which then decays to a beta particle and an antineutrino. The details
36
about bosons will not be discussed now.
D → U + W–
W–
→
e0 +  e
–1
(2) Mesons are unstable particles. Pions and other mesons are present in atmosphere
formed by the interaction cosmic rays with it. Pions decay into leptons and their neutrinos as
follows. There are two modes of the decay of charged pions.
Muon emission:
      
     
Electron emission:
   e  e
   e   e
Neutral pion decays by the following two ways
 0  2
 0  e  e  
GLUONS: Gluons are chargeless and massless force carrier particles which bind the quarks
in forming the mesons and baryons.
FERMIONS: Particles containing odd number of elementary particles called are spin half
particles(do not bother about this term spin half now). Examples are quarks, leptons, baryons.
BOSONS: Particles having even number elementary particles are called spin 1 particles(forget
about spin 1 now). Mesons are bosons as they contain two quarks. Force carrier particles like
photons, gluons, W and Z bosons are also included in this category.
FORCE CARRIER PARTICLES:
Photons, gluons, W and Z bosons are called force carrier particles.
LARGE HADRON COLLIDER(LHC):
The mimicry of the big bang is being tried 100 m below ground near Geneva with a 27
Km(circumference) circular gigantic tunnel in the France-Switzerland border surrounding Alps
mountain. The protons(handron) are allowed to collide with each other in opposite directions
at a velocity close that of light so as to produce elementary particles and their antiparticles,
bosons, gluons and other particles. Thus it will recreate conditions that existed during 1st
billionth of a second after the big bang. Scientists from more than 100 nations are working for
the mega project day and night to prove or disprove the big bang theory of creation of universe.
The most important particle they expect to find is HIGG'S BOSON. Higg's boson is the
hypthetical particle which was responsible, as they believe, to give mass to massless photons.
After more than two LHC Collisions by now, many new findings have been obtained. But
scientists are not sure of the presence of Higg's boson yet. If not found, then the big bang
37
theory will go wrong. Some other theory will come up for the creation of the urniverse. The
other particles they expect are SUPPER SYMMETRIC PARTICLES. which compose of
96% of universe called DARK MATTER. Astronauts have got the evidence of the existence
of dark matter from the unusually high gravitational pull they experience which is not matching
with the material universe.
LHC experiment consists of the following steps.
1. Linear Accelerator : The hydrogen gas is subjected to electron impact to break it
into atoms and strip off one electron from each to covert them to protons. These protons then
are allowed to enter the linear accelerator in which they are subjected to an electric field to get
velocity nearly one third of light.
2. Four Ring Circular Boosters: The protons are then enter into four circular boosters
in which they are subjected to electric and magnetic field. Here they are compressed and get
velocity 91.6% of light. Each ring has 157 m circumference.
3. Proton Synchroton: Here again it is subjected to magnetic fields to increase its
velocity to 99.9% of light. Note that when a particle travels with velocity nearly equal to that of
light, if we give more energy to it, its relativistic mass will increase instead of velocity. Here
protons become 25 times heavier than its rest mass. Proton synchroton is circular in shape
having circumference of 628 m.
4. Super Proton Synchroton : Again protons are stronger magnetic fields to acquire
more energy. Its energy becomes 450 GeV(Giga electron volt). Its mass increase much more.
It is also circular in shape having 7 Km circumference.
5 Large Hadron Collider : Ultimately the protons enter into LHC in two counter
rotating beams, but not allowed to collide. They are subjected to more energy to acqure a
value of 7 TeV(tera electron volt). The proton becomes 7000 times heavier than its rest mass
here and rotate 11000 times a second. LHC has 27 Km circumference. After gaining the
required energy here they are allowed to collide at 4 positions(Dectector) where the elementary
particles, their antiparticles, bosons, gluons, super symmetric particles, Higg's boson etc. are
expected to be detected.
Time will tell us what is the outcome of the mamoth scientific endeavour.
SAQ 17: Can you say, an antiproton is made up of what elementary particles ? How is
antihydrogen atom is composed of ?
RESPONSE TO SAQs
Radioactivity
SAQ 1:
(i)
Th230 --------> 2He4 + 88Ra226 (ii)
Th232--------> 2He4 + 88Ra228
90
90
Both the isotopes of Th i.e 230 and 232 are alpha emitters. Both produce the same elements no
doubt(Ra) but they are different isotopes of Ra, one is Ra-226 and the other Ra-228 isotope.
(iii)
Ra226 --------> 2He4 + 86Rn222
(iv) 86Rn220 --------> 2He4 + 84Po216
88
38
SAQ 2:
Th234 --------> -1e0 + 91Pa234
(i)
90
We found in the previous SAQ that Th-232 and Th-230 isotopes are alpha emitters while Th-234
we find here is a beta emitter.
(ii)
K39 -------> -1e0 + 20Ca39
(iii)91Pa234 -------> -1e0 + 92U234
19
60
0
60
(iv)
Co -------> -1e + 28Ni
(v) 83Bi212 -------> -1e0 + 84Po212
27
238
SAQ 3: 92U is an alpha emitter and its daughter 90Th234 is a beta emitter. We have seen these
examples before. There is no relationship between the nature of the emission of parent and
daughter nuclides. The parent nuclide can be an alpha emitter while the daughter might be a beta
emitter. In some other cases both parent and daughter nuclides can be either alpha or beta
emitters.
SAQ 4:
(i)
Mg27 ---------> -1e0 + 13Al27
(ii)
Bi210 --------> -1e0 + 84Po210
12
83
213
4
209
(iii)
Po ---------> 2He + 82Pb
(iv)
O15---------> +1e0 + 7N15
84
8
(v)
Au200--------> -1e0 + 80Hg200
(vi)
Br82--------> -1e0 + 36Kr82
79
35
226
4
222
(vii)
Ra --------> 2He + 86Rn
(viii)
Po210-------> 2He4 + 82Pb206
88
84
38
0
38
(ix)
K ---------> +1e + 18Ar
(x)
I122 --------> +1e0 + 52Te122
19
53
(xi)
Fe60 --------> -1e0 + 27Co60
(xii)
Np230-----> 2He4 + 91Pa226
26
93
46
0
46
221
(xiii)
V ---------> 1e + 22Ti
(xiv)87Fr
-----> 2He4 + 85At217
23
SAQ 5: Stable nuclides which fall inside the band: Cl-37, K-39, Ne-20, C-13
Unstable and radioactive nuclides: Mg-27: lies above the band(because Mg-24 is stable)
P-30: lies below the band(because P-31 is stable), K-40: lies above the band(as K-39 is stable)
O-15: lies below the band(as O-16, 17 and 18 are stable),
B-10: lies below the band(as
B-11 is stable), Br-82: lies above the band(as Br-79 and Br-81 are stable); Fe-60:lies above the
band(as Fe-56 is stable); Ag-111: lies above the band(as Ag-108 is stable).
SAQ 6: These are beta emitters because their n/p ratio are more than expected(usual). The
stable nuclides of these elements are 1H1, 1H2, 35Br79 and 35Br81, 26Fe56, 26Br79,Br81 and 47Ag108.
Since the given nuclides have higher mass numbers, they have more number of neutrons. They
fall above the band of stability. They get themselves stabilised by emitting a beta particle.
H3 --------> -1e0 + 2He3
Br82 -------->-1e0 + 36Kr82,
1
35
Fe60 ---------> -1e0 + 27Co60
Co60 --------> -1e0 + 28Ni60.
27
111
0
111
Ag --------> -1e + 48Cd .
47
SAQ 7: These are positron emitters because, their n/p ratio are less than expected(usual). The
stable nuclides of these elements are, P-31, N-14, Ne-20, Na-23, Sc-45, S-32, V-51, Mn-55. But
the mass numbers given are less than their respective stable values. So they are positron emitters.
By emitting positron, the n/p ratio increases in the daughter nuclides and become more stable.
P28 ---------> +1e0 + 14Si28,
N12 ----------> +1e0 + 6C12
15
7
18
0
18
Ne --------> +1e + 9F ,
Na21 --------> +1e0 + 10Ne21
10
11
41
0
40
Sc --------> +1e + 20Ca ,
S30 ----------> +1e0 + 15P30
21
16
46
0
46
V ---------> +1e + 22Ti ,
Mn50 --------> +1e0 + 24Cr50
25
23
SAQ 8: Beta emitters: 6C14, 11Na24, 19K40, 12Mg27, 25Mn56, 53I129, 26Co60
Their mass numbers are greater than those of their stable mass numbers given in the SAQ.
Positron emitters: 5B10, 11Na22, 19K37, 15P28, 21Sc41,
26
39
Their mass numbers are less than those of thier stable mass number values given in the SAQ.
Write the products in each case for practice.
SAQ 9:
(i)
For converting 10gm to 5gm(half), one half life period is required. Again for converting
5gms to 2.5gm, another half life period is required. So for the conversion of 10gms of C-14 to
2.5gms of C-14, two half life periods i.e 2X 5730 = 11460 years will be required.
(ii)
For converting 1gm to 1/2 gm, we need one half life, then for converting 1/2 to 1/4gm we
need another half life, then for converting 1/4 gm to 1/8 gm we need one more half life and finally
for converting 1/8 to 1/16 gm, we need one more half life period. So a total of four half life
periods i.e 4 X 4.5 X 109 years will be required for converting 1 gms of U-238 to 1/16gm.
Alternatively:
At = A0 1
2
⇒
t
T0 .5
1/16 = 1 X (1/2)x (where x = t/T0.5= number of half lives)
⇒ (1/2)4 = (1/2)x
⇒
x =4
(four half lives)
So the time required = 4 X 4.4 X 109 years.
Which method you liked, the second, the formula method or the first, analysis method?
(iii)
Let us use the formula to get the answer quickly in this case.
At = A0 1
2
t
T0 .5
⇒ At = 12.8(1/2)7 = 12.8 X 1/128 = 0.1 gm
(since t/T0.5=number of half life periods)
So after 7 half life periods, 12.8gms of tritium will be reduced to only 0.1 gm.(Note that this
method is simpler than analysis method, where you have to find the fraction left after 7 half lives
and then multiply this fraction with 12.8gms to get the answer).
SAQ 10: This would be solved in the same way as shown in the example. The solution is left to
you which you can easily do after learning logarithm chapter in mathematics.
log(12) = log(15.3) + (t/5730) log (1/2),
⇒ t = 2012 years
SAQ 11:
(i)
N 14: No. of protons =7, No. of neutrons =7 and No. of electrons=7
7
Calculated atomic mass = 7 X 1.0073 + 7X 1.0087 + 7X 0.00055 = 14.11585amu
Actual isotopic mass = 14.0067amu(given), So Δm = 14.11585 - 14.0067 = 0.10915amu
So binding energy(B.E) of 7N14 = 0.10915 X 931 = 101.62 MeV( Since 1 amu= 931MeV)
Binding energy per nucleon(B.E per nucleon) = 101.62/14= 7.26 MeV.
(ii)
P31: No. of protons = 15, No. of neutrons = 16, No. of electrons= 15
15
Calculated atomic mass= 15 X 1.0073 + 16X1.0087 + 15 X 0.00055 = 31.25695amu
Actual isotopic mass = 30.9740amu, So Δm = 31.25695-30.9740 =0.28295amu
So binding energy(B.E) = 0.28295 X 931 = 263.42 MeV
Binding energy per nucleon(B.E per nucleon) = 263.42/31 = 8.497 MeV.
If we compare between the binding energy per nucleon(B.E per nucleon) between the two
nuclides, then we find that 15P31 nuclide has greater B.E per nucleon than 7N14 nuclide.
SAQ 12:
There are 150 pairs (300 isotopes) belonging to 37 elements those are obtained
during the fission of U-235. Some pairs are given in the text.
40
235
92
U
In these reactions, two neutrons are produced along with the daughter nuclides.
SAQ 13:
Fission reaction produces much greater energy as the mass loss is more.
SAQ 14:
(ii) 13Al27 + 2He4 -----> 0n1 + 15P30
(i)7N14 + 2He4 --------> 1H1 + 8O17
(iii) 3Li7 + 1H1 ---------> 0n1 + 4Be7
(iv) 83Bi209 + 1H2 -------> 0n1 + 84Po210
SAQ 15:
(ii) 1H2 (deuteron)
(iii) -1e0(electron)
(iv)
(i) 26Fe53
2
Na
11
(v) 1H1
SAQ 16:
N14 + 0n1 -------> 1H1 + 6C14**(radioactive)
7
14
Since C has more neutrons than expected(C12 and C13 are stable), it is a beta emitter.
C14 --------> -1e0 + 7N14
6
This is the process that takes place in the carbon dating which we have discussed before.
SAQ 16 : Let x number of alpha and y number of beta particles are emitted in the uranium
series. Let us build up two simultaneous equations from mass numbers and atomic numbers
from LHS and RHS.
92
U238 =
82
Pb206 + x 2He4 + y -1e0
238 = 206 + 4x + 0
(1)
92 = 82 + 2x - y
(2)
On solving, x = 8 and y = 6
You can get these numbers in the same way for other series.
SAQ 17 : Antiproton is made up of two U (anti U) quarks and one D (anti D) quark. So what
is the total charge ? Is it not –1 (-2/3-2/3+1/3).
In an antihydrogen atom, the neucleus is made up of an antiproton, already discussed and
antielectron(positron) would be rotating about the nucleus
IMPORTANT: According to recent convention, the correct
representation of a nuclide has both the mass number and atomic
number written in the LHS as follows.
235
92
U
So the readers are advised to write according to the above format
and not put the mass numbers in the RHS subscript as given in
the entire chapter. For saving time, this has been done in this
book. However it is not according to recent convention. Please
bear with this inconvenience.
41