Half-Life Example: radon (nasty stuff*) has a half-life of 3.8
days. If you start with 1 mg of radon, after 3.8 days you will
have 0.5 mg of radon.
Upcoming Schedule
Season’s
Greetings!
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please!
Dec. 1
10.6-10.7
Dec. 3
10.6
Quiz 10
Dec. 5
10.7
Dec. 8
11.1-11.4,
11.7
Dec. 10
12.1-12.3
Reactor Tour
Dec. 12
12.4-12.12
Final Exam
Thursday, December 18
10:30 a.m.
“What we say, what we do in each individual case, may move the whole world.
And that puts an exceptional responsibility on our shoulders.”—Edward Teller
radiometric dating
Carbon-14 dating is the best-known example. Carbon-14 is
formed in the atmosphere by the reaction
14
7
N + 01 n → 146 C + 11 H .
This reaction is continually taking place in the atmosphere, and
the carbon-14 atoms are continually beta decaying to nitrogen14, with a half-life of 5760 years.
Because carbon-14 is continually being created and decaying,
we eventually reach a steady state condition, where there is a
constant amount of carbon-14 in the atmosphere.
Days
0
3.8
7.4
11.4
Radon Left (mg)
1
0.5
0.25
0.125
The mean lifetime of a nucleus is different than its half-life. It
turns out that
T = 1.44 T1/2 .
See Beiser for details but don’t worry about this for the final.
*But perhaps not as dangerous as once believed.
Living things take up carbon-14 as long as they are alive, and
have the same ratio of carbon-14 to carbon-12 as does the
atmosphere.
When living things die, they stop taking up carbon-14, and the
radioactive carbon-14 decays.
If we compare the carbon-14 to carbon-12 ratio in a dead
organism with a living one, we can tell how long the carbon-14
has been decaying without replenishing, and therefore how
long the organism has been dead.
This assumes he carbon-14 to carbon-12 ratio in the
atmosphere is the same now as it was when the
organism died.
It also assumes living organisms now are essentially the
same in their carbon content as were similar organisms
long ago.
1
Carbon-14 dating takes us back a relatively short time, and
both assumptions seem to be valid.
The formula for radiocarbon dating, derived from R = R0 e-λt, is
t =
R
1
ln 0 .
λ
R
A similar approach can be taken with radioactive potassium,
rubidium, or uranium, to go back much further in time.
We have to find parent-daughter decay schemes that give us
unique daughter nuclei; i.e., they could have only come from
decay of the parent.
We need to know the activity R0 of the organism at death,
which is the reason for the second assumption on the previous
slide.
If we assume the daughter nuclei came only from the original
radioactive nuclei, we can calculate the original number, and
then calculate the decay time.
Radiocarbon dating is good for a few half-lives of carbon-14, or
50,000 or so years.
We measure the time back to some event caused the clock to
start "ticking;" i.e., an event that froze into the sample the
particular number of parent atoms which resulted in the
observed number of daughter atoms.
Radiocarbon example. A piece of wood has 13 disintegrations
per minute per gram of carbon. The activity of living wood is 16
dpm per gram. How long ago did the tree die?
t =
t =
R
1
ln 0 .
λ
R
5760 years
16
ln
= 1726 years .
0.693
13
12.3 Radioactive Series
FS2003—skip this section (good idea to read it anyway).
Most radionuclides belong to one of four radioactive series.
12.4 Alpha Decay
Alpha decay happens when a nucleus is too
large for the short-range forces between
nucleons to hold it together.
Consider a parent nucleus, which emits an alpha particle, and
changes into a daughter nucleus.
The initial mass is just the mass of the parent.
The final mass is the mass of the daughter + alpha. In
the case of alpha decay, this final mass must be less
than the parent mass.
The energy released is the mass difference times c2. This
is the energy available to get the alpha particle out of
the nucleus.
2
Q = ( mi - mf - mα ) c 2
Alpha particle binding energy:
is the energy released in the decay.
Why does an alpha particle escape from the nucleus, rather
than individual neutrons and protons?
An alpha particle's mass is considerably smaller than the
sum of the masses of its constituents. (This is another
way of saying the alpha particle has a high binding
energy.) We need this big mass difference to provide the
energy to get the alpha out.
If calculate Q for neutrons or protons, you don't end up
with enough available energy to get them out of the
nucleus. In fact, you need to put energy in.
So an alpha particle can escape the nucleus, and mass is
converted to energy in the process.
This is "like" an explosion. Both the alpha particle and the
daughter nucleus will recoil, in opposite directions.
The kinetic energy carried away by the alpha particle is
approximately
A-4
A is parent nuclide
Kα =
Q.
mass number.
A
The mass difference, converted to MeV, is about 28 MeV
(huge!).
Hey! You said always include the electrons. Where are they?
To include the electrons, include them in the “mass of parts”
and use “mass of helium” instead of “mass of alpha.” You get
the same result.
Example:
222.
86Rn
222
86
218
82
Po + 42 He
Notice how the total electron mass is included on both sides of the reaction.
Q = ( mradon - mpolonium - mα ) c 2
 931MeV 
Q = ( 222.017574 u - 218.008930 u - 4.002603 u ) 

u


Q = 5.587 MeV
This equation comes from the conservation of energy and
momentum (an easy Physics 23 calculation).
Most alpha emitters are very massive compared to the alpha
particle, so the alpha particle's KE is nearly equal to the total
energy of the disintegration (Q).
Rn →
Kα =
Kα =
A-4
Q
A
222 - 4
Q = 5.486 MeV
222
3
Alpha particles escape the nuclear potential by tunneling out. I
won’t test you on the tunnel theory of alpha decay.
12.5 Beta Decay
Beta decay occurs when a nucleus has too many neutrons
relative to protons. Emission of an electron by a neutron
changes it to a proton.
The electron immediately leaves the nucleus upon creation. Of
course, this leaves the atom in an ionized state, but it will
quickly find an electron somewhere and become neutral.
Also, beta decay involves n → p + e-, but spins of all three
particles are 1/2. There is no way (in the above “reaction”) for
the net spin to be the same before and after; i.e., angular
momentum appears not to be conserved.
Finally, conservation of linear momentum would require
electrons and nuclei to recoil in opposite directions, but this is
also not observed experimentally.
The solution to these problems is simple in hindsight; another
particle must be involved in beta decay.
The "extra" particle is the neutrino. It is not charged and has
no detectable mass (if it does have mass, the mass is very
small).
Again, there are some questions. The energy spectrum of
emitted electrons typically looks like this:
The beta particles come
out with less energy than
the Q of the reaction.
http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/beta.html
It is logical to claim that electrons are created with the
maximum energy, and lose some of it to the nucleus on the
way out. But this is not verified experimentally. So, what
happened to the missing energy?
Lacking charge and mass, the neutrino interacts very weakly
with matter, which is why it took so long to discover.
Beta decay is, then,
n → p+ + e - + ν .
actually, an antineutrino (that’s
what the bar means)
The above “reaction” shows you what took place but cannot be
used for energy calculations; you must include the parent and
all daughter products.
Electron capture and positron emission, listed separately from
beta decay at the beginning of this lecture, are actually just
variations on beta decay.
4
Positron emission is like beta decay, except a +e particle is
emitted. In positron emission, a proton is converted into a
neutron, a positron, and, of course, a neutrino:
p+ → n+ e+ + ν .
Competitive with positron emission is electron capture:
p+ + e- → n+ ν .
We found the “strong” force is required to explain how protons
in nuclei can be held together.
The “weak” force—the fourth and final (?) force in nature, is
needed to explain radioactivity.
There is another nuclear reaction by which a proton can change
into a neutron. It is called inverse beta decay:
p+ + ν → n+ e+ .
Technically, this is not radioactive decay because it involves an
antineutrino from outside “colliding” with a nucleus.
Inverse beta decay was used to confirm the existence of
neutrinos.
The absorbed electron usually comes from the K shell, and later
on an electron from an outer shell drops into the empty state,
emitting a gamma ray.
Electron capture occurs more often than positron emission in
heavy elements because the inner electrons are close to the
nucleus.
12.6 Gamma Decay
No final exam questions from this section. The basic
idea is that a too-energetic nucleus gets rid of energy
by emitting a high-energy photon.
12.7 Cross Sections
Good stuff. Not covered Fall 2003.
Another kind of inverse beta decay is
n+ ν → p+ + e - .
Both kinds of inverse beta decay have extremely low
probabilities, which is why we aren't bothered by the billions of
neutrinos passing through us right now.
5
12.8 Nuclear Reactions
Example:
14
7
3
1
Here's how to think of nuclear reactions:
An energetic projectile strikes a target nucleus.*
If the projectile energy is large enough, or if it is a
neutral particle like a neutron, the projectile can reach
and stick to the target.
The result is a new, compound nucleus, which is
probably unstable.
After a relatively short period of time, the compound
nucleus decays, leaving a new nucleus. Energy and
radiation are given off.
N* (figure 12.15).
H + 116 N → 147 N* → 136 C + 11 H .
This is just one of many possible nuclear reactions involving
nitrogen-14.
If you plot the probability of the above reaction occuring vs.
energy of the incident tritium, you might something like figure
12-14 in Beiser, or the crude figure below:
prob.
this is called a “resonance”
the resonance has an
energy width
*Those of you who toured the reactor learned that the energy of the
neutron which causes U-235 to fission is actually very small, and
energetic neutrons actually cause no fission.
The energy width of the resonance can be used to define the
lifetime of the resonant state ( 147 N*).
Heisenberg's uncertainty principle says
=
∆E ∆t ≥
.
2
If we consider the width of the resonance to be ∆E/2, often
called Γ, and the lifetime τ to be the uncertainty in time ∆t,
then the excited state lifetime is
=
τ =
.
Γ
This is how we measure lifetimes of excited nuclear states.
Center of Mass Coordinate System -- skip this subsection. You
will not be tested on it.
energy
12.9 Nuclear Fission
We have done most of the groundwork for
understanding this section. Nuclear fission is just a
nuclear reaction. Here’s a famous example:
235
92
140
U + 01 n → 236
92 U* → 54 Xe +
94
38
Sr + 01 n + 01 n.
Several hundred MeV of energy, mostly in the form of
kinetic energy of the fission fragments, is also released.
Know how to calculate that energy! (It’s like a binding
energy problem.)
Other reactions are possible. See Beiser figure 12.20.
Other numbers of neutrons are released in other
reactions.
6
Here are the important points:
Lots of energy is released by a single fission.*
Each fission of uranium-235 produces, on the average,
2.5 neutrons
If, on the average, one of those 2.5 neutrons can be
used to produce another fission, you have a selfsustaining chain reaction.
How do you sustain a chain reaction?
More than 1.5 neutrons per fission escaping to the
“outside” will stop a chain reaction.
To sustain a chain reaction, you "trap" the neutrons
(e.g., surround the fissioning material with a neutron
reflector, such as beryllium).
Or you make the fissioning mass physically bigger in
extent, so that neutrons have more matter to pass
through before they escape. “Critical mass.”
To make a bomb, “all”* you have to do is assemble a mass of
fissionable material greater than the critical mass (a few
kilograms).
*Be able to calculate the energy!
Tickling the dragon’s tail.
One of the major tasks of those designing the first atomic bomb
was to determine the critical mass of the plutonium to be used.
One can make calculations, but experiment is the only way to
make sure your calculations are correct.
An experiment was devised in which
a neutron source was placed inside a
hollow plutonium sphere. The
plutonium alone was sub-critical.
The neutron source would help kick-start the chain reaction, but
the assembly was still subcritical with the neutron source inside.
*Acquiring the fissionable material is a whole different story. Also, in designing your
bomb, you might wish to allow enough time for yourself to vacate the area.
The assembly could be made critical by placing neutron
reflectors around the sphere. The reflectors would reflect
neutrons back into the plutonium.
Tungsten-carbide “bricks” were used
as reflectors.
Harry Daghlian was working alone one
evening on the experiment, adding WC bricks and listening to his geiger
counters for evidence of criticality.
7
As Daghlian lifted by hand the final
brick over the assembly, the increased
clicking of the geiger counter warned
him that the assembly was approaching
criticality.
I have read that “tickling the dragon’s tail” was a highlight of
lab tours sometimes given to distinguished visitors.
As he withdrew the final brick with his left hand, it slipped out
of his hand and dropped on top of the plutonium sphere.
Moving quickly, he inserted his right hand into the blue glow
surrounding the plutonium sphere and pushed the fatal brick
aside.
Daghlian died 26 days later as a result of massive exposure to
radiation. A guard seated a few meters away survived a mild
case of radiation sickness.
After World War II, criticality experiments continued.
On May 21, 1946, Louis Slotin (who had been a friend of
Daghlian) was instructing colleagues in a modified experiment,
in which two sub-critical plutonium hemispheres were brought
together until a critical mass was approached.
Slotin held the top hemisphere in one hand, and slowly lowered
it towards the bottom one. A screwdriver in his right hand kept
the two hemispheres from touching…
…until it slipped out of Slotin’s grasp. The top hemisphere fell
on top of the bottom one.
It was claimed that the
room filled with a blue
glow, and a wave of heat
surged over the
observers.
Ignoring the intense
burning pain in his left
hand, Slotin knocked the
two hemispheres apart.
Nine days later, Slotin died of radiation sickness. The others
survived various degrees of illness, although three of them died
years later from complications that might have been caused by
the accident.
Why wasn’t the apparatus designed with a fixed upper
hemisphere and a lower one to be raised into place?
8
Here are some fusion reactions which might be useful in fusion
reactors:
12.10 Nuclear Reactors
Out of time! If there is time, I will lecture briefly on this section
from memory.
12.11 Nuclear Fusion in Stars
2
1
2
1
3
1
12.12 Fusion Reactors
I won’t test you specifically on material from sections 12.10
through 12.12, but you are able to calculate energies released
in fission and fusion reactions. Therefore, be prepared for test
problems on such calculations!
H + 21 H → 31 H + 11 H + 4.0 MeV
H + 21 H → 32 He + 01 n + 3.3 MeV
H + 21 H → 42 He + 01 n +17.6 MeV
The last reaction is particularly attractive because it liberates so
much energy.
Know how to calculate the energies!
Advantages of fusion:
Abundant materials. Deuterium is readily available and
tritium is easy to produce.
Clean. "No" nasty radioactive fission products. Only
hydrogen, helium, neutrons, energy. (Tritium, however,
is radioactive, but you can “burn” it.)
Disdvantages of fusion:
“just” an
engineering
problem
Extremely high energies needed to begin a reaction.
How do you "hold" the "burning" fuel. (You don’t.)
How do you manage to bring a material object close
enough to the burning fuel to extract energy without
vaporizing the material?
9