The mean lives of a radioactive substance are 1620 years and 405 years for `alpha`-emission and `beta`-emission, respectively. Find out the time during which three fourths of a sample will decay if it is decaying both by `alpha`-emission and `beta`-emission simultaneously.

Explanation

The decay constant `lamda` is the reciprocal of the mean life `tau `
Thus, `lamda_(alpha)=1/(1620)` per year and
`lamda_(beta) = 1/(405) ` per year
`:.` Total decay constant , `lamda=lamda_(alpha)+lamda_(beta)` (or)
`lamda=-(1)/(1620)+1/(405)=1/324`
per year . We know that `N=N_(0)e^(-lamdat)`
When `3/4` th part of the sample has disintegrated ,
`N = N_(0)//4`
`:.N_(0)/4=N_(0)e^(-lamdat)("or")e^(lamdat)=4`
Taking logarithm of both sides , we get
`lamdat = log_(e)4` (or)
`t=(1)/lamdalog_(e)2^(2) =2/lamdalog_(e) 2 = 449` year