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Question

Three-fourth of the active nuclei present in a radioactive sample decay in 34 s. The half-life of the sample is

A
38 s
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B
34 s
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C
12 s
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D
1 s
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Solution

The correct option is A 38 s
Given that, at t=34 s, (34)th of the sample decayed, then the fraction of remaining nuclei is,

NN0=14

NN0=14=12n [NN0=12n]

2n=4=22

n=2

Using, t=nT1/2

34=2T1/2

T1/2=38 s

Hence, option (A) is correct.
Why this question?

To understand the decay law and apply it to solve the problem related to half life period.

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