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The half-life (T(1//2) ) of a radioactiv...

The half-life `(T_(1//2)` ) of a radioactive substance is 60 min, after 3 h how much percentage will left of this substance?

A

50

B

75

C

25

D

`12.5`

Text Solution

Verified by Experts

The correct Answer is:
D
Number of half-life `(t )/( T _(1//2))= 3 xx 60` min
`= (3 xx 60)/(60) = 3`
` therefore N = N _(0) ((1)/(2)) ^(t //T _(1//2))`
`N = N _(0) ((1)/(2)) ^(3)`
`N = N _(0) ((1)/(8))`
`implies ( N )/( N _(0)) xx 100 = 100 xx (1)/(8)`
`= 12.5 %`
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