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Radon-222 is a radioactive gas for which...

Radon-222 is a radioactive gas for which the decay constant is `2.1xx10^(-6)s^(-1)`. If the initial decay rate is `5.6xx10^(10)s^(-1)`, calculate :
(i) the initial number of radioactive atoms present and
(ii) the time which will elapse before the activity is reduced to one-quarter of its initial value. Given : `log_(e)2=0.693`.

Text Solution

Verified by Experts

(i) Initial decay rate = `lamdaN_(0)`
`5.6xx10^(10)=2.1xx10^(-6)N_(0)`
or `N_(0)=(5.6xx10^(10))/(2.1xx10^(-6))=2.67xx10^(16)`
(ii) Activity at any time t is proportional to number of atoms present at time t.
`N=N_(0)/4`
Now, `N_(0)/4=N_(0)e^(-lamdat)ore^(-lamdat)=4`
Taking logs, `lamdat=log4=2log_(e)2=2xx0.693`
`t=(2xx0.693)/lamda=(2xx0.693)/(2.1xx10^(-6))s`
= `0.66xx10^(6)s=6.6xx10^(5)s`
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