I.E. Irodov Solutions on Radioactivity

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.

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?

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 Na²⁴ radionuclide whose half-life is 15 hours?

Here

Also,

So the number of β rays emitted in one hour is No (1 – e^-λt) = 1.13 x10¹⁵

4. To investigate the beta-decay of Mg²³ radionuclide, a counter was activated at the moment t = 0. It registered N1 beta-particles by a moment t₁ = 2.0s, and by a moment t₂ = 3t₁ the number of registered beta-particles was 2.66 times greater. Find the mean lifetime of the given nuclei.

If N₀ 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.

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?

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 Co⁵⁵ radionuclide if its activity is known to decrease 4.0% per hour. The decay product is nonradioactive.

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 U²³⁸ preparation of mass 1.0 g emits 1.24 X10⁴ alpha particles per second. Find the half-life of this nuclide and the activity of the preparation.

Here

= 2.531 X 1021

The activity is A = 1.24 x 10⁴ dis/sec

Hence the half life is

9. Determine the age of ancient wooden items if it is known that the specific activity of C¹⁴ nuclide in them amounts to 3/5 of that in lately felled trees. The half-life of C¹⁴ nuclei is 5570 years.

In old wooden atoms, the number of C¹⁴ nuclei steadily decreases because of radioactive decay. (In live trees biological processes keep replenishing C¹⁴ nuclei maintaining a balance. This balance starts getting disrupted as soon as the tree is felled.)

If T1/2 is the half life of C¹⁴ then

Hence,

10. In a uranium ore the ratio of U²³⁸ nuclei to Pb²⁰⁶ nuclei is η = 2.8. Evaluate the age of the ore, assuming all the lead Pb²⁰⁶ to be a final decay product of the uranium series. The half-life of U²³⁸ nuclei is 4.5.109 years.

What this implies is that in the time since the ore was formed

11. Calculate the specific activities of Na²⁴ and U²³⁵ nuclides whose half-lifes are 15 hours and 7.1 x 10⁸ years respectively.

The specific activity of Na²⁴ is

Here M = molar weight of N_a²⁴= 24 g N_A is Avogadro number & T_1/2 is the half-life of Na²⁴ Similarly the specific activity of U²³⁵ is

= 0.793 x 10⁵dis/ (g.S)

12. A small amount of solution containing Na²⁴ radionuclide with activity A = 2.0 x 10³disintegrations per second was injected in the bloodstream of a man. The activity of 1 cm³of blood sample taken t = 5.0 hours later turned out to be A’ = 16 disintegrations per minute per cm³. The half-life of the radionuclide is T = 15 hours. Find the volume of the man’s blood.

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 Na²⁴ 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 10³]/[16/60] x e ^-ln(⅔)cc = 5.99 litre

13. The specific activity of a preparation consisting of radioactive Co⁵⁸ and nonradioactive Co⁵⁹ is equal to 2.2 x10¹² dis/(s•g). The half-life of Co⁵⁸ 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)

We see that Specific activity of the sample

=1/[M+M’]

(Activity of M gm of Co⁵⁸ 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 Co⁵⁸.

= M/58 x 6.023 x 10²³ x ln 2/[71.3 x 86400] dis/sec

From the given data,

1.168 x 10¹⁵[M/(M+M’)] = 2.2 x 10¹²

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:

Find the half-lifes of both components and the ratio of radioactive nuclei of these components at the moment t = 0.

Suppose N₁ N₂ are the initial number of component nuclei whose decay constants are λ₁ , λ₂ ( 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 λ₂ = 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 λ₁ = 0.66 per hour

T₁ = 1.05 hour and T₂ = 9.9 hours [half-lives]

Ratio: N₁/N₂ = [51.1/10.0] x λ₂/ λ₁ = 0.54

15. A P³² radionuclide with half-life T = 14.3 days is produced in a reactor at a constant rate q = 2.7×10⁹ nuclei per second. How soon after the beginning of production of that radionuclide will its activity be equal to A = 1.0 x 10⁹ dis/s?

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 A₁ with decay constant λ₁ transforms into a radionuclide A₂ with decay constant λ₂. Assuming that at the initial moment the preparation contained only the radionuclide A₁, find:

(a) the equation describing accumulation of the radionuclide A₂ With time;

(b) the time interval after which the activity of radionuclide A₂ reaches the maximum value.

(a) Suppose N₁ and N₂ are the number of two radionuclides A₁, A₂ at time t. Then

(b) The activity of nuclide A₂ is λ₂ N₂. This is maximum when N₂ is maximum. That happens when

dN₂/dt = 0

This requires

λ₂ e^{-λ_2 t_m} = λ₁ e^{-λ_1 t_m}

t_m = [ln(λ₁/λ₂)/(λ₁-λ₂)

17. Solve the foregoing problem if λ₁ = X ₂ = X.

(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 A₁ goes through the transformation chain A₁ → A₂ → A₃ (stable) with respective decay constants λ₁ and λ₂. Assuming that at the initial moment the preparation contained only the radionuclide A₁ equal in quantity to N₁₀ nuclei, find the equation describing accumulation of the stable isotope A₃.

Here we have the equations

dN₁/dt = -λ₁N₁ and dN₂/dt = -λ₂N₂ and dN₃/dt = -λ₃N₃

we from previous problem, N₁ = N₁₀ e^-λ_1t

19. A Bi²¹⁰ radionuclide decays via the chain

where the decay constants are λ₁= 1.60 X 10^-6 S^-1, λ₂= 5.80×10^-8 s^-1. Calculate alpha- and beta-activities of the Bi²¹⁰ preparation of mass 1.00 mg a month after its manufacture.

We have the chain

of the previous problem initially

A month after preparation

N₁ = 4.54 x 10¹⁶

N₂ = 2.52 x 10¹⁸ using the results of the previous problem.

Then

Aβ = λ₁N₁ = 0.725 x 10¹¹ dis/sec

A_α = λ₂N₂ = 1.46 x 10¹¹ dis/sec

20. (a) What isotope is produced from the alpha-radioactive Ra²²⁸ 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 Pb²⁰⁶ isotope?

(a) Ra has Z – 88, A – 226 After 3 a emission and 4 p (electron) emission

A – 206

Z – 88 + 4-5×2 – 82

The product is ⁸²Pb₂₀₆

(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