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

The decay constant for the first process is `lambda_(1)=(ln 2)/(t_(1))` and for the second process it is `lambda_(2)=(ln)/(t_(1))`. The probability that an active nucleus decays by the first process in a time interval `dt` is `lambda_(1)dt`. Similarly, the probability that it decays by the second process is `lambda_(2)dt`. The probability that it either decays by the first process or by the second process is `lambda_(1)dt+lambda_(2)dt`. If the effective decay constant is `lambda`, this probability is also equal to `lambda dt`. Thus.
`lambda dt=lambda_(1)dt+lambda_(2)dt`
or , `lambda-lambda_(1)+lambda_(2)`
or `(1)/(t)=(1)/(t_(1))+(1)/(t_(2))` To be remembered