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the count rate from a radioactive sample...

the count rate from a radioactive sample falls from `4.0xx10^6`per second to `1.0xx10^6` per second in 20 hours.What will be the count rate 100 hours after the begnning ?

Text Solution

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If `A_(0)` is the initial activity of radioactive sample then acitivity at any time
`A=A_(0)e^(-lamda t)`
or `1xx10^(6)=4xx10^(6)e^(-lamda)xx20`
or `e^(-20lamda)=1/4`
Thecount rate after 100 hour is given by
`A.=A_(0)e^(-lamdaxx100)=A_(0)e^(-100lamda)`
`=A_(0)[e^(20lamda)]^(5)`
`=4xx10^(6)[1/4]^(5)`
`=3.91xx10^(3)` per second
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