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

Explanation

The decay constant `lambda` is the reciprocal of the mean life, `tau`
Thus, `lambda_alpha=1/1620` Per year and `lambda_beta=1/405` Per year
`therefore` Total decay constant , `lambda=lambda_alpha+lambda_beta`
`lambda=1/1620+1/405=1/324` year
We know that `N=N_0 e^(-lambdat)`
When `3/4`th part of the sample has disintegrated,
`N=N_0//4 therefore N_0/4 = N_0e^(-lambdat)` or `e^(lambdat)`=4
Taking logarithm on both sides , we get `lambdat=log_e 4`
`t=1/lambdalog_e 2^2 = 2/lambda log_e2`
=2 x 324 x 0.693 = 449 years.