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

Decay constant for the first process is
`lambda_(1)=0.693/(t_(1))`,
and for the second process `lambda_(2)=0.693/(t_(2))`
The probability that an active nucleus decays by the first process in small time dt is `lambda_(1)dt`. Similarly, the probability that it decays either by first process or by second process is `lambda_(2) dt`. The probability that it decays either by first process or by second process is `(lambda_(1)dt+lambda_(2)dt)`. If effective decay constant is `lamdba`, this probability is also equal to `lambda dt`.
`:. lambda dt=lambda_(1)dt+lambda_(2)dt`
`lambda=lambda_(1)+lambda_(2)`
`0.693/t=0.693/(t_(1))+0.693/(t_(2)) :. 1/t=1/(t_(1))+1/(t_(2))`
which was to be proved.