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The t(1//2) of radioactive K-40 is 5.274...

The `t_(1//2)` of radioactive K-40 is 5.274 years `(lambda = 2.5 xx 10^(-7) "min"^(-1))`. The decay activity of 2.0 g of the sample is about

A

`5 xx 10^(5)` dpm

B

`5 xx 10^(10)` dpm

C

`7.5 xx 10^(15)` dpm

D

`7.5 xx 10^(20)` dpm

Text Solution

Verified by Experts

The correct Answer is:
C
`lambda=2.5xx10^(-7) "min"^(-1)`
Mass of `K*40=2*0 g`
`N=(2.0)/(40)xx6.02xx10^(23)`
`=3.01xx10^(22)`
Rate of decay `=lambdaN`
`=2.5xx10^(-7)xx3.01xx10^(22)`
`=7.5xx10^(15)` dpm
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DINESH PUBLICATION-NUCLEAR AND RADIO CHEMISTRY-Evaluate Yourself
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