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The fraction of a radioactive material w...

The fraction of a radioactive material which remains active after time t is `9//16`. The fraction which remains active after time `t//2` will be .

A

`4/5`

B

`3/5`

C

`3/4`

D

`7/8`

Text Solution

Verified by Experts

The correct Answer is:
C
First order decay
`N(t)=N_(0)e^(-lambdat)`
Given `N(t)//N_(0) = 9//16 = e^(-lambdat)`
Now , N(t/2) = `N_(0)e^(-lambdat//2)`
`(N(t//2))/(N_(0))=sqrt(e^(-lambdat))=sqrt(9//16)`
N(t/2) = `3//4N_(0)` .
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