A radioactive nucleus can decay by two different processes. The half-life for the first process is `t_1` and that for the second process is `t_2`. Show that the effective half-life `t` of the nucleus is given by
`1/t = 1/t_1 + 1/t_2`.

Explanation

Let N be the total number of atoms of the radioactive sample initially. Let `(dN_1)/(dt)` and `(dN_2)/(dt)`
be the initial rates of disintegrations of the radioactive sample by the two processes respectively. Then
`(dN_1)/(dt)=lambda_1N` and `(dN_2)/(dt)=lambda_2N`
where `lambda_1` and `lambda_2` are the decay constants for the first and second processes respectively.
The initial rate of disintegrations of the radioactive sample by both the processes
`=(dN_1)/(dt)+(dN_2)/(dt)=lambda_1N+lambda_2N=(lambda_1+lambda_2)N`
If `lambda` is the effective decay constant of the radioactive sample, its initial rate of disintegration.
`(dN)/(dt)=lambdaN`, But `(dN)/(dt)=(dN_1)/(dt)+(dN_2)/(dt)`
`lambdaN=(lambda_1+lambda_2)N , lambda=lambda_1+lambda_2`
`0.693/T_1+0.693/T_2=0.693/T , 1/T=1/T_1+1/T_2 ,T=(T_1T_2)/(T_1+T_2)`