The document Radioactivity (Part - 1) - Atomic and Nuclear Physics, Irodov JEE Notes | EduRev is a part of the JEE Course I. E. Irodov Solutions for Physics Class 11 & Class 12.

All you need of JEE at this link: JEE

**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 Na ^{24 }radionuclide whose half-life is 15 hours? **

**Ans. **Here

= 2.51 x 10^{16}

Also

So the number of β rays emitted in one hour is

No (1 - e^{-λt}) = 1.13 x10^{15}

**Q.217. To investigate the beta-decay of Mg ^{23} radionuclide, a counter was activated at the moment t = 0. It registered N_{1} 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.**

**Ans. ** If N_{0} 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 Co ^{55 }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 U ^{238 }preparation of mass 1.0 g emits 1.24.10^{4} alphaparticles per second. Find the half-life of this nuclide and the activity of the preparation.**

**Ans.** Here

= 2.531 X 10^{21 }

The activity is A = 1.24 x 10^{4} dis/sec .

Then 4.90 x 10^{18} per sec .

Hence the half life is

4.49 x10^{9} years

**Q.222. 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. **

**Ans. **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 T_{1/2} is the half life of C^{14} then

Hence = 4105 years ≈ 4.1x10^{3 }years

**Q.223. 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 4.5.10^{9 }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 10^{9} years.

**Q.224. Calculate the specific activities of Na ^{24} and U^{235 } nuclides whose half-lifes are 15 hours and 7.1.10^{8 }years respectively.**

**Ans. **The specific activity of Na^{24} is

= 3.22 x 10^{17 }dis/(gm.sec)

Here M = molar w eight of is Avogadro num ber & is the half-life of Na^{24}

Sim ilarly the specific activity of U^{235} is

= 0.793 x 10^{5 }dis/(gm-s)

**Q.225. A small amount of solution containing Na ^{24} radionuclide with activity A = 2.0.10^{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. **

**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 Na^{24} 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 Co ^{58} and nonradioactive Co^{59} is equal to 2.2.10^{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 per cent). **

**Ans.** We see that

Specific activity of the sample

{A ctivity of M gm of Co^{58} in the sample}

Here M and At are the masses of Co^{58} and Co^{59} in the sample. Now activity of M gm of Co^{58}

= 1.168 x l0^{15}M

Thus from the problem

= 2.2 x 10^{12}

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 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 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 P ^{32 } radionuclide with half-life T = 14.3 days is produced in a reactor at a constant rate q = 2.7.10^{9} nuclei per second. How soon after the beginning of production of that radionuclide will its activity be equal to A = 1.0.10^{9} 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 A _{1} with decay constant λ_{1} transforms into a radionuclide A_{2} with decay constant λ_{2}. Assuming that at the **

(a) the equation describing accumulation of the radionuclide A

(b) the time interval after which the activity of radionuclide A

**Ans. **(a) Suppose N_{1} and N_{2 }are the number of two radionuclides A_{1}, A_{2} at time t. Then

(1)

(2)

From (1)

where N_{10} is the initial number of nuclides A_{l} at time t = 0

From (2)

or

since N_{2 }= 0 at t = 0

Constant

Thus

(b) The activity of nuclide A_{2} is λ_{2} N_{2}. This is maximum when N_{2} is maximum. That happens when

This requires

or

**Q.230. Solve the foregoing problem if λ _{1} = X_{2 }= 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

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!