Lecture 4
Problem Set-1
Objectives
In this lecture you will learn the following
In this lecture, we shall practice solving problems.We will solve 5 representative problems.
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Lecture 4
Problem Set-1
Background Information
Mole – Molecular weight expressed in grams.
1 mole of any element has Avogadro number of molecules. Every molecule has 1 nucleus.
1 MeV = 1.603 x 10-13 Joules.
Avogadro Number = 6.023 x 10 23.
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Lecture 4
Problem Set-1
Question-1
How may neutrons and protons are there in the nuclei of the following atoms; (a) Li7, (b) Mg 24,(c) Xe 135, (d) Rn 222.
Sol. The following website gives properties of isotopes.
http://ie.lbl.gov/education/isotopes.htm
The Berkeley Laboratory Isotopes Project
Isotopes of Lithium (Z=3)Click on an isotope to get more
information about its decay
Isotope
Half-life
6Li
stable
7Li
stable
8Li
838 ms
9Li
178.3 ms
Now from the above source or from any book we can construct the following table.
Mass
Number
7
Protons/ Electrons
3
Neutrons
Li
Charge
Number
3
Mg
12
24
12
12
4
Xe
54
135
54
81
Rn
86
222
86
136
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Lecture 4
Problem Set-1
Question-2
Tritium ( H3 ) decays by negative beta decay with a half-life of 12.33 years. The atomic weight of H3
is 3.016. What is the mass in grams of 1 mCi of tritium ?
Solution - 2
Activity = α = 1 mCi = 3.7 1010 x 10-3 = 3.7 x 107 s-1 = λ N
T1/2 = 12.33 years
λ = 0.693/T 1/2
=0.693/(12.33 x 365.25 x 24 x 3600)
= 1.781 x 10-9 s-1
N= 3.7 x 107/1.781 x 10-9
= 2.08 x 1016 nuclei
Mass of Tritium Atom = 3.016/6.023 x 1023
= 5.007 x 10-24 g
Hence total mass =5.007 x 10-24 x 2.08 x 1016
= 1.040 x 10-7 g
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Lecture 4
Problem Set-1
Question 3
The radioisotope battery is fuelled with 500g of Pu238C (Plutonium-238 carbide), which has a density
of 12.5 g/cm3. The Pu238 has a half life of 89 years, and emits 5.6 MeV per disintegration, all of
which may be assumed to be absorbed in the generator. The thermal to electrical efficiency of the
system is 6 percent. Calculate (a) the specific power in watts (thermal) per gram of fuel; (b) the
power density in watts (thermal) per cm 3; (c) the fuel efficiency in curies per watt (thermal); (d) the
total electrical power of the generator.
Solution 3
Mol-wt-PuC
250
T 1/2
89
years
Energy per disintegration
5.6
MeV
Density - PuC
12.5
g/cm3
Total Mass of PuC
500
g
Efficiency
0.06
λ = 0.693/T 1/2
= 0.693/(89 x 365.25 x 24 x 3600)
= 2.467 x 10-10 s-1
Molecules/unit mass = N = 6.023 x 1023/250
= 2.409 x 1021 g-1
Specific activity = λN = 2.467 x 10-10 x 2.409 x 1021
= 5.944 x 1011 s-1/g
= 5.944 x 1011 /3.7 x 1010
= 16.07 Ci/g
Specific Power= λNE = 5.944 x 1011 x 5.6 x 1.603 x 10-13
= 0.534 W/g
Power Density = λNEρ = 0.534 W/g x 12.5 g/cm3
= 6.67 W/cm 3 Fuel Efficiency = λN/ λNE = 16.07 Ci/g / 0.534 W/g
= 30.11 Ci/W Total Power = Specific power x Mass x Conv. Eff.
=λnEmη = 0.534 W/g x 500 g x 0.06
= 16.01 W(electric) 1
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Lecture 4
Problem Set-1
Question 4
Three elements A, B and C, not connected by a chain, decay individually. At t=10s, the ratio of their
activities are 1:0.5:0.25, and at t=15s, their activities are in the ratio 1:0.25:0.0625. What would be the
ratio of their activities at t=0s.
Solution 4
Similarly
From above, we can
write,
Using the data at 10 s,
(1)
Using the data at 15 s,
Dividing the latter by former we get,
Substituting the value in Eq. (1), we get
Since the ratio of αB/ αC is similar, we can write,
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Lecture 4
Problem Set-1
Question 5
Tata Memorial Hospital has been procuring a short-lived radio-isotope for medical applications. The
doctors, of late, had been complaining that the supplied isotope did not have the necessary strength
for which the order was placed. In order to verify this, the inspection department carried out the
following test. The sample was counted for 1 minute for the number of disintegration, immediately
on its arrival. In the first one minute 1791 disintegrations were counted. After 10 minutes from the
arrival time, the sample was again counted for 1 minute. This time the number of counts were 1620.
Based on the above facts, compute the strength of the source in Becquerels (dps), when supplied.
Solution 5
Every disintegration results in a radiation and this can be detected in a counter. Thus if a counter
detects N counts, it implies N disintegrations have taken place.
Let the Number of nuclei in the sample at t=0 be N(0)
Number of Nuclei after one minute
=
This implies that
(1)
Similarly,
(2)
Dividing Eq. (1) by Eq. (2), we get
Substituting this in Eq. (1), we get
Activity at t = 0 is given by
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