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Q.

The radioactivity of an element becomes $\frac{1}{64} t h$ of its original value in 60 sec. Then half life period is

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a

10 sec

b

30 sec

c

5 sec

d

20 sec

answer is B.

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Detailed Solution

use $A = \frac{A_{0}}{64} = \frac{A_{0}}{2^{6}} s o 6 h a l f l i f e p e r i o d = 60 s e c 1 h a l f l i f e = 10 s e c$ .

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