Irodov Solutions: Radioactivity- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE PDF Download

Q.214. 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. 

Ans. (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 

Therefore the mean life time is

Q.215. What fraction of the radioactive cobalt nuclei whose halflife is 71.3 days decays during a month? 

Ans. We calculate λ first

Hence

fraction decaying in a month 

Q.216. How many beta-particles are emitted during one hour by 1.0μg of Na²⁴  radionuclide whose half-life is 15 hours? 

Ans. Here 

= 2.51 x 10¹⁶

Also 

So the number of β rays emitted in one hour is

No (1 - e^-λt) = 1.13 x10¹⁵

Q.217. 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.

Ans.  If N₀ is the number of radionuclei present initially, then

where Then

or  

Substituting the values

Put Then

or 

Now 

Negative sign has to be rejected as x >0.

Thus   x = 0.882

This gives        

Q.218. The activity of a certain preparation decreases 2.5 times after 7.0 days. Find its half-life.

Ans. If the half-life is T days

Hence 

or  

Q.219. 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? 

Ans.  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 nucli decreases by a factor

So activity decreases to articles per minute.

Q.220. 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. 

Ans.  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

or   = 0.0408 per hour

he mean life time is 24.5 hour

Q.221. A U²³⁸  preparation of mass 1.0 g emits 1.24.10⁴ alphaparticles per second. Find the half-life of this nuclide and the activity of the preparation.

Ans.  Here

= 2.531 X 10²¹ 

The activity is                  A = 1.24 x 10⁴ dis/sec .

Then   4.90 x 10¹⁸ per sec .

Hence the half life is

 4.49 x10⁹ years

Q.222. 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. 

Ans. 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 T_1/2 is the half life of C¹⁴ then  

Hence = 4105 years ≈ 4.1x10³ years

Q.223. 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.

Ans. What this implies is that in the time since the ore was formed   nuclei have remained undecayed. Thus

or

Substituting  years, η = 2.8

we get              t = 1.98 x 10⁹ years.

Q.224. Calculate the specific activities of Na²⁴  and U²³⁵  nuclides whose half-lifes are 15 hours and 7.1.10⁸ years respectively.

Ans. The specific activity of Na²⁴ is

 = 3.22 x 10¹⁷ dis/(gm.sec)

Here M = molar w eight of  is Avogadro num ber &  is the half-life of  Na²⁴

Sim ilarly the specific activity of U²³⁵ is

= 0.793 x 10⁵ dis/(gm-s)
 

Q.225. A small amount of solution containing Na²⁴  radionuclide with activity A = 2.0.10³ disintegrations per second was injected in the bloodstream of a man. The activity of 1 cm³ of 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. 

Ans. 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

Now   t = 5 hour

Thus  = 5.99 litre

Q.226. The specific activity of a preparation consisting of radioactive Co⁵⁸  and nonradioactive Co⁵⁹ is equal to 2.2.10¹² 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 per cent). 

Ans. We see that

Specific activity of the sample 

 {A ctivity of M gm of Co⁵⁸ in the sample} 

Here M and At are the masses of Co⁵⁸ and Co⁵⁹ in the sample. Now activity of M gm of Co⁵⁸

= 1.168 x l0¹⁵M

Thus from the problem

 = 2.2 x 10¹²

or  

Q.227. 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. 

Ans. 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 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  in the data for small t ]

Thus we get  λ_{1 =} 0.66 per hour

Ratio 

The answer given in the book is misleading.
 

Q.228. 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 = 1.0.10⁹  dis/s? 

Ans. Production of the nucleus is governed by the equation

We see that N will approach a constant value   This can also be proved directly. Multiply by  and write

Then

or 

At t = 0 when the production is starteed, N = 0

Hence

Now the activity is 

From the problem

This gives λt = 0.463 

so 

Algebraically       

Q.229. 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. 

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

             (1)

        (2)

From (1)

where N₁₀ is the initial number of nuclides A_l at time t = 0

From (2)

or  

since         N₂ = 0 at t = 0

Constant 

Thus 

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

This requires

or

Q.230. Solve the foregoing problem if λ₁  = X₂ = X.

Ans. a) This case can be obtained from the previous one on putting

where ε is very small and letting ε → 0 at the end. Then

or dropping the subscript 1 as the two values are equal

(b) This is maximum when