The following Question appeared in Karnataka CET 2003 question paper:

80 = 100/2

20 = 100/2

where ‘n’ and ‘m’ are the numbers of half lives required for 20% decay and 80% decay respectively.

Dividing, 80/20 = 2

Now consider the following MCQ which appeared in Karnataka CET 2004:

**Half life of a radioactive substance is 20 min. The time between 20% and 80% decay will be**

(a) 25 min. (b) 30 min. (c) 40 min. (d) 20 min

If the initial activity is A0, the activity ‘A’ after ‘n’ half lives is given by

A = A

Let us take the initial activity as 100 units. After 20% decay, the activity becomes 80 units and after 80% decay, the activity becomes 20 units. These two cases can be stated as(a) 25 min. (b) 30 min. (c) 40 min. (d) 20 min

If the initial activity is A0, the activity ‘A’ after ‘n’ half lives is given by

A = A

_{0}/2^{n}.80 = 100/2

^{n}and20 = 100/2

^{m}where ‘n’ and ‘m’ are the numbers of half lives required for 20% decay and 80% decay respectively.

Dividing, 80/20 = 2

^{m}/2^{n}= 2^{(m–n)}. Or, 2^{(m-n)}= 4, from which (m–n) = 2 half lives = 2×20 min. = 40 min.Now consider the following MCQ which appeared in Karnataka CET 2004:

**A count rate meter shows a count of 240 per minute from a given radioactive source. One hour later the meter shows a count rate of 30 per minute. The half life of the source is**

(a) 80 min. (b) 120 min. (c) 20 min. (d) 30 min.

From the equation, A = A(a) 80 min. (b) 120 min. (c) 20 min. (d) 30 min.

_{0}/2^{n}, we have 30 = 240/2^{n}so that n = 3. Therefore one hour is equal to 3 half lives which means the half life of the substance is 20 min.
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