Two Kerala Engineering Entrance 2007 Questions from Nuclear Physics

The following MCQ (on radioactivity) which appeared in Kerala Engineering Entrance 2007 question paper is simple, but it differs slightly from the conventional type:

Radium has half life of 5 years. The probability of decay of a radium nucleus in 10 years is

(a) 50% (b) 75% (c) 100% (d) 60% (e) 25%

Since the half life is 5 years, the amount getting decayed in 10 years will be 75%.

[This can be found very easily: After 5 years half the initial amount will be decayed; after another 5 years, half of the remaining will be decayed. If the half life and the time period given are not so simply related, you will have to calculate the number of nuclei undecayed (N) using the equation, N = N₀/2ⁿ where N₀ is the initial number and ‘n’ is the number of half lives in the given time. The percentage decayed in the given time is then calculated].

The probability for decay of any given nucleus in 10 years is therefore 75%.

The following question involving the relative abundance of isotopes is a popular one and it has found place in Kerala Engineering Entrance 2007 question paper:

The natural boron of atomic weight 10.81 is found to have two isotopes B¹⁰ and B¹¹. The ratio of abundance of isotopes in natural boron should be

(a) 11:10(b) 81:19 (c) 10:11 (d) 15:16 (e) 19:81

This is a simple arithmetical problem. If there are n₁ boron atoms of atomic weight 10 and n₂ boron atom of atomic weight 11 in a sample of natural boron, the mean atomic weight (which is given as 10.81) is related to n₁ and n₂ as

(10n₁ + 11n₂)/ (n₁ + n₂) = 10.81.

Rearranging, 0.81n₁ = 0.19n₂, from which n₁/n₂ = 19/81.

Radioactive Decay Law

The radioactive decay law as you might be remembering well is expressed mathematically as
N = N0e^-λt with usual notations.
In most entrance examinations such as Medical and Engineering entrance examination, you wont be allowed to use calculators or logarithm tables. The above equation, modified in terms of half life will be very useful in this context. If N is the number of nuclei remaining undecayed after ‘n’ half life periods, it is related to the initial number N0 as,
N = N0/2ⁿ.
Now let us discus the following M.C.Q.:
Out of 1.414×10²⁴ nuclei, only 10²⁴ nuclei remain undecayed after 15 minutes in a radioactive sample. The half life period of the sample in minutes is
(a) 64 (b) 55 (c) 40 (d)30 (e) 24
We have, 10²⁴ = (1.414×10²⁴)/2ⁿ from which 2ⁿ = 1.414 so that n= ½. This means that 15 minutes is half of the half life period. The half life of the sample therefore is 30 minutes.
The above question can be asked in a modified manner, involving the activity of the sample as follows:
The activity of a radioactive sample drops from 1.414×10⁸ disintegrations per second to 10⁸ disintegrations per second in 15 minutes. The half life period of the sample in minutes is
(a) 64 (b) 55 (c) 40 (d)30 (e) 24

Since the activity of a sample is directly proportional to the number of nuclei present at the instant, we can express the activity ‘A’ after ‘n’ half lives in terms of the initial activity ‘A0’ as,
A = A0/2ⁿ
Substituting the values of A and A0, we have 10⁸ = (1.414×10⁸)/2ⁿ from which n=½. So 15 minutes is half of the half life period of the sample and the answer to the question is 30 minutes [option (d)].