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The ratio activity of an element becomes...

The ratio activity of an element becomes `1//64 th` of its original value in `60 sec`. Then the half-life period is

A

5 sec s

B

10 sec s

C

20 sec s

D

30 sec s

Text Solution

Verified by Experts

The correct Answer is:
B
`2^(n)=64 rArr n=6 therefore T_(1//2)=(60)/(6)=10s`
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