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_(1)N` and `(dN_(2))/(dt)=lambda_(2)N`
Where `lambda_(1)` and `lambda_(2)` are the decay constants for the first and second processes respectively.
The initial rate of disintergrations of the radioactive sample by both the processes
`=(dN_(1))/(dt)+(dN_(2))/(dt)=lambda_(1)N+lambda_(2)N=(lambda_(1)+lambda_(2))N`.
If `lambda` is the effective decay constant of the redioactive sample, its initial rate of disintergration.
`(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_(1)T_(2))/(T_(1)+T_(2))`