Radioactivity is a phenomenon exhibited by a few matters of emitting energy and subatomic particles spontaneously. Radioactivity is due to the nuclear instability of an atom. This chapter deals with the properties of radioactivity of an atom. IE IRODOV Physics Solutions provides the solutions to various problems related to the concept. The questions are diversified covering all types and aid in understanding the basic fundamentals of the concept easily.
The solutions are based on the specific formulae, equations for every question. They help the students understand the topic and connect the theoretical and practical problems. Solving these kinds of problems helps to have a grip over the chapter. This makes the student confident and gives a good practice for the upcoming board and competitive examination. Solving numerous problems and examples enable the students to develop an interest in the subject and it automatically motivates them to aim for higher grades.
1. Knowing the decay constant λ of a nucleus, find:
(a) the probability of decay of the nucleus during the time from 0 to t;
(b) the mean lifetime ζ of the nucleus.
Solution:
a) The probability of survival (i.e. not decaying) in time t is e^{-λt}. Hence the probability of decay is 1 - e^{-λt}
b) The probability that the particle decays in time dt around time t is the difference
e^{-λt} - e^{-λ(t+dt)} = e^{-λt} [1 - e^{-λt dt }] = λe^{-λt} dt
Therefore, the mean life-time is
= 1/λ
2. What fraction of the radioactive cobalt nuclei whose halflife is 71.3 days decays during a month?
Solution:
At first, we will calculate λ
Hence fraction decaying in a month = 1-e^{λt}= 0.253
3. How many beta-particles are emitted during one hour by 1.0μg of Na^{24} radionuclide whose half-life is 15 hours?
Solution:
Here
Also,
So the number of β rays emitted in one hour is No (1 - e^{-λt}) = 1.13 x10^{15}
4. To investigate the beta-decay of Mg^{23} radionuclide, a counter was activated at the moment t = 0. It registered N1 beta-particles by a moment t_{1} = 2.0s, and by a moment t_{2} = 3t_{1} the number of registered beta-particles was 2.66 times greater. Find the mean lifetime of the given nuclei.
Solution:
If N_{0} is the number of radio nuclei present initially, then
The negative sign has to be rejected as x >0.
Thus x = 0.882
5. The activity of a certain preparation decreases 2.5 times after 7.0 days. Find its half-life.
Solution:
If the half-life is T days
Hence,
6. At the initial moment, the activity of a certain radionuclide totalled 650 particles per minute. What will be the activity of the preparation after half its half-life period?
Solution:
The activity is proportional to the number of parent nuclei (assuming that the daughter is not radioactive). In half its half-life period, the number of parent nuclei decreases by a factor.
So activity decreases to
7. Find the decay constant and the mean lifetime of Co^{55 }radionuclide if its activity is known to decrease 4.0% per hour. The decay product is nonradioactive.
Solution:
If the decay constant (in (hour)-1 ) is λ., then the activity after one hour will decrease by a factor e~λ
Hence 0.96 = e~λ
λ = 1.11 X 10^{-5} S^{-1}= 0.0408 per hour
The mean lifetime is 24.5 hour
8. A U^{238} preparation of mass 1.0 g emits 1.24 X10^{4} alpha particles per second. Find the half-life of this nuclide and the activity of the preparation.
Solution:
Here
= 2.531 X 1021
The activity is A = 1.24 x 10^{4} dis/sec
Hence the half life is
9. Determine the age of ancient wooden items if it is known that the specific activity of C^{14 }nuclide in them amounts to 3/5 of that in lately felled trees. The half-life of C^{14 }nuclei is 5570 years.
Solution:
In old wooden atoms, the number of C^{14} nuclei steadily decreases because of radioactive decay. (In live trees biological processes keep replenishing C^{14 }nuclei maintaining a balance. This balance starts getting disrupted as soon as the tree is felled.)
If T1/2 is the half life of C^{14 }then
Hence,
10. In a uranium ore the ratio of U^{238 }nuclei to Pb^{206 }nuclei is η = 2.8. Evaluate the age of the ore, assuming all the lead Pb^{206} to be a final decay product of the uranium series. The half-life of U^{238} nuclei is 4.5.109 years.
Solution:
What this implies is that in the time since the ore was formed
11. Calculate the specific activities of Na^{24 }and U^{235 }nuclides whose half-lifes are 15 hours and 7.1 x 10^{8} years respectively.
Solution:
The specific activity of Na^{24} is
Here M = molar weight of N_{a}^{24}= 24 g N_{A }is Avogadro number & T_{1/2} is the half-life of Na^{24 }Similarly the specific activity of U^{235} is
= 0.793 x 10^{5}dis/ (g.S)
12. A small amount of solution containing Na^{24 }radionuclide with activity A = 2.0 x 10^{3}disintegrations per second was injected in the bloodstream of a man. The activity of 1 cm^{3}of blood sample taken t = 5.0 hours later turned out to be A' = 16 disintegrations per minute per cm^{3}. The half-life of the radionuclide is T = 15 hours. Find the volume of the man's blood.
Solution:
Let V = volume of blood in the body of the human being. Then the total activity of the blood is A' V. Assuming all this activity is due to the injected Na^{24} and taking account of the decay of this radionuclide, we get
VA’ = A e^{-λt}
Now, λ = (ln 2)/15 per hour and t = 5 hour
Therefore, V = (A/A’) e^{- ln(⅔) }= [2.0 x 10^{3}]/[16/60] x e^{ -ln(⅔)}cc = 5.99 litre
13. The specific activity of a preparation consisting of radioactive Co^{58 }and nonradioactive Co^{59} is equal to 2.2 x10^{12} dis/(s•g). The half-life of Co^{58} is 71.3 days. Find the ratio of the mass of radioactive cobalt in that preparation to the total mass of the preparation (in percent)
Solution:
We see that Specific activity of the sample
=1/[M+M’]
(Activity of M gm of Co^{58} in the sample)
Here M and At are the masses of Cobalt 58 and Cobalt 59 in the sample. Now activity of M gm of Co^{58}.
= M/58 x 6.023 x 10^{23} x ln 2/[71.3 x 86400] dis/sec
From the given data,
1.168 x 10^{15}[M/(M+M’)] = 2.2 x 10^{12}
Or M/(M+M’) = 1.88 x 10^{-3}.
14. A certain preparation includes two beta-active components with different half-lifes. The measurements resulted in the following dependence of the natural logarithm of preparation activity on time t expressed in hours:
t | 0 | 1 | 2 | 3 | 5 | 7 | 20 | 14 | 20 |
ln A | 4.10 | 3.60 | 3.10 | 2.60 | 2.06 | 1.82 | 1.60 | 1.32 | 0.90 |
Find the half-lifes of both components and the ratio of radioactive nuclei of these components at the moment t = 0.
Solution:
Suppose N_{1} N_{2 }are the initial number of component nuclei whose decay constants are λ_{1} , λ_{2 }( in (hour)^{-1} Then the activity at any instant is
The activity so defined is in units dis/hour. We assume that data In A given is of its natural logarithm. The daughter nuclei are assumed nonradioactive.
We see from the data that at large t the change in In A per hour of elapsed time is constant and equal to - 0.07.
Thus λ_{2} = 0.07 per hour We can then see that the best fit to data is obtained by
[To get the fit we calculate We see that A(t)=e^{0.07 t} it reaches the constant value 10.0 at t = 7, 10, 14, 20 very nearly. This fixes the second term. The first term is then obtained by subtracting out the constant value 10.0 from each value of A(t)=e^{0.07 t} in the data for small t ]
Thus we get λ_{1} = 0.66 per hour
T_{1} = 1.05 hour and T_{2} = 9.9 hours [half-lives]
Ratio: N_{1}/N_{2} = [51.1/10.0] x λ_{2}/ λ_{1} = 0.54
15. A P^{32} radionuclide with half-life T = 14.3 days is produced in a reactor at a constant rate q = 2.7x10^{9} nuclei per second. How soon after the beginning of production of that radionuclide will its activity be equal to A = 1.0 x 10^{9} dis/s?
Solution:
Production of the nucleus is governed by the equation
dN/dt = g - λN
Where g -> supply and λN -> decay
We observe that N will approach a constant value g/λ. This can also be proved directly. Multiply by and write
At t = 0, when the production is started, N=0
so, 0 = g/λ + constant
Hence, N = (g/λ) (1 - e^{-λt})
Now the activity is A = λN = g(1 - e^{-λt})
Also from the problem, we know that
1/2.7 = 1 - e^{-λt}
This gives λt = 0.463
So, t = [0.463xT)/0.693 = 9.5 days
Algebraically, t = -T/(ln 2) ln (1 - A/g)
16. A radionuclide A_{1} with decay constant λ_{1} transforms into a radionuclide A_{2} with decay constant λ_{2}. Assuming that at the initial moment the preparation contained only the radionuclide A_{1}, find:
(a) the equation describing accumulation of the radionuclide A_{2} With time;
(b) the time interval after which the activity of radionuclide A_{2} reaches the maximum value.
Solution:
(a) Suppose N_{1} and N_{2} are the number of two radionuclides A_{1}, A_{2} at time t. Then
(b) The activity of nuclide A_{2} is λ_{2} N_{2}. This is maximum when N_{2} is maximum. That happens when
dN_{2}/dt = 0
This requires
λ_{2} e^{-λ_2 t_m } = λ_{1} e^{-λ_1 t_m}
t_{m} = [ln(λ_{1}/λ_{2})/(λ_{1}-λ_{2})
17. Solve the foregoing problem if λ_{1} = X _{2} = X.
Solution:
(a)This case can be obtained from the previous one by putting
Where ε is very small and ε ->0 at the end. Then
Or dropping the subscript at 1 as the two values are equal, we get
(b) This is maximum when
18. A radionuclide A_{1} goes through the transformation chain A_{1} → A_{2} → A_{3 } (stable) with respective decay constants λ_{1} and λ_{2}. Assuming that at the initial moment the preparation contained only the radionuclide A_{1} equal in quantity to N_{10} nuclei, find the equation describing accumulation of the stable isotope A_{3}.
Solution:
Here we have the equations
dN_{1}/dt = -λ_{1}N_{1} and dN_{2}/dt = -λ_{2}N_{2 } and dN_{3}/dt = -λ_{3}N_{3}
we from previous problem, N_{1} = N_{10} e^{-λ_1t}
19. A Bi^{210} radionuclide decays via the chain
where the decay constants are λ_{1}= 1.60 X 10^{-6}_{S-1, λ2= 5.80x10-8 s-1. Calculate alpha- and beta-activities of the Bi210 preparation of mass 1.00 mg a month after its manufacture. }
Solution:
We have the chain
of the previous problem initially
A month after preparation
N_{1} = 4.54 x 10^{16}
N_{2} = 2.52 x 10^{18 } using the results of the previous problem.
Then
Aβ = λ_{1}N_{1} = 0.725 x 10^{11} dis/sec
A_{α} = λ_{2}N_{2} = 1.46 x 10^{11} dis/sec
20. (a) What isotope is produced from the alpha-radioactive Ra^{228} as a result of five alpha-disintegrations and four β-disintegrations?
(b) How many alpha- and β-decays does U ^{238} experience before turning finally into the stable Pb^{206} isotope?
Solution:
(a) Ra has Z - 88, A - 226 After 3 a emission and 4 p (electron) emission
A - 206
Z - 88 + 4-5x2 - 82
The product is ^{82}Pb_{206}
(b) We require,
-ΔZ = 10 = 2n - m
-ΔA = 32 = ft x 4
Here n = no. of α emissions
m = no. of β emissions
Thus n = 8, m = 6