The mean lives of an unstable nucleus in two different decay processes are `1620 yr` and `405 yr`, respectively. Find out the time during which three-fourth of a sample will decay.

Explanation

Radioactivity is the statistical phenomenon. In statistics the important theorem is that the probabilities of individual and independent events. Let `lambda_(alpha)` and `lambda_(B)` be the decay constant for `alpha` -and `beta `-emission, respectively. According to theorem, the decay constant for composite event is `lambda_(alpha) + lambda_(beta)`.
If T is half-life of composite event and `T_(alpha)` and `T_(beta)` are the half -lives of `alpha -and beta-emisiion`, then `lambda =(0.693)/(T)`
`(1)/(T) =(1)/(T_(alpha)) +(1)/(T_(beta))` or `(1)/(tau) =(1)/(tau_(alpha) _ (1)/(tau_(bet))a)`
this gives
`tau = tau_(apha) (tau_(beta))/(tau_(alpha))+(tau_(beta=1620 xx 405))/(1620 + 405) =324` years
Let t be the time in which the given sample decays three - fourth. Therefore, the fraction of sample undeecayed in time t is`1//4`. That is,
`(N)/(N_(0)) =(1)/(4)` From the relation `N=N_(0) e^(-lambda t)`
or `log_(e) =(N)/(N_(0)) = - lambda t`
`rArr t =(1)/(lambda) lag_(e) (N_(0))/(N) =tau log_(e) (N_(0))/(N)`
`=2.3026 tau log_(10) (N_(0))/(N)`
`=2.3026 xx 324 xx 0.6031`
Required time, `t=449.94` years .