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Question

A sample of radioactive material A, having an activity of 10 mCi (1 Ci=3.7×1010 decays/sec) has twice the number of nuclei as another sample of a different radioactive material B, which has an activity of 20 mCi. The correct choices for half-lives of A and B would be, respectively,

A
20 days and 5 days
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B
10 days and 20 days
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C
5 days and 10 days
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D
20 days and 10 days
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Solution

The correct option is A 20 days and 5 days
The activity is given by,
A=λN

For substance A:-

AA=λANA 10=NAλA

As, NA=2NB

10=2NBλA ......(i)



For B:-

AB=NBλB

20=NBλB ......(ii)

Dividing (i) by (ii), we get,

1020=2NBλANBλB

λB=4λA

Since, T1λ

TATB=λBλA=4λAλA

TA=4TB

Considering all the options, the only option that satisfies the above condition is TB=5 days and TA=20 days

Hence, (D) is the correct answer.

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