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 fourth of a sample will decay if it is decaying both by `alpha`-emission and `beta`-emission simultaneously. `(log_e4=1.386).`

Explanation

Here, `tau_(alpha)=1620`years, `tau_(beta)=405"years"`.
`t=? N=N_0-3/4N_0=(N_0)/4`
If `lambda_(alpha)` and `lambda_(beta)` are decay constant for alpha and
beta emission respectively,
then, `lambda_(alpha)=1/(tau_(alpha))=1/1620yr^-1`
`lambda_(beta)=1/(tau_(beta))=1/405yr^-1`
Total decay constant `lambda=lambda_(alpha)+lambda_(beta)`
`=1/1620+1/405=(1+4)/1620`
`=1/324yr^-1`
form `N=N_0e^(-lambda t)`,
`1/4N_0=N_0e^(-lambda t)`
`t=(log_e4)/lambda=1.386/(1//324)yr`
`= 449.1 yr`