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

Explanation

The decay constant `lamda` is 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 on both sides , we get
`lamdat = log_(e) 4 " or " t = 1/lamdalog_(e) 2^(2) = 2/lamdalog_(e)2`
`= 2 xx 324 xx 0.693 = 449` year .