A radioactive sample decays in two modes. In one mode its half life is ${t_1}$ and in the other mode its half life is ${t_2}$. Find the overall half life
(A) ${t_1} + {t_2}$
(B) $\dfrac{{{t_1} + {t_2}}}{2}$
(C) $\dfrac{{{t_1}{t_2}}}{{{t_1} + {t_2}}}$
(D) $\dfrac{{{t_1}{t_2}}}{{{t_1} - {t_2}}}$

Hint: Half life is defined as the number of atoms reduced to half of its present value.
${t_{^{\dfrac{1}{2}}}} = \dfrac{{\log {\,_e}\,2}}{\lambda }$

Complete step by step answer:
${\lambda _1} = \dfrac{{{{\log }_e}2}}{{{t_1}}}$ …………. (i)
${\lambda _2} = \dfrac{{{{\log }_e}2}}{{{t_2}}}$ ………… (ii)
Where ${t_1}$ and ${t_2}$ are their respective half life and ${\lambda _1}$ and ${\lambda _2}$ are decay constants.
So the decay rate of quantity $N$ is given by
$ - \dfrac{{dN}}{{dt}} = N{\lambda _1} + N{\lambda _2}$
$ \implies \dfrac{{dN}}{N} = - ({\lambda _1} + {\lambda _2})dt$ …………. (iii)
Integrate equation (iii)
$\int_{{N_o}}^N {\dfrac{{dN}}{N}} = - \int_o^t {({\lambda _1} + {\lambda _2})\,dt} $
$ \implies \left| {{{\log }_e}N} \right|_{{N_o}}^N = - ({\lambda _1} + {\lambda _2})[t]_o^t$
$ \implies {\log _e}N - {\log _e}{N_o} = - ({\lambda _1} + {\lambda _2})t$
$ \implies {\log _e}\dfrac{N}{{{N_o}}} = - \left( {{\lambda _1} + {\lambda _2}} \right)t$
Take antilog on both sided
$\dfrac{N}{{{N_o}}} = {e^{ - ({\lambda _1} + {\lambda _2})t}}$
$ \implies N = {N_o}{e^{ - ({\lambda _1} + {\lambda _2})t}}$ …………. (iv)
Use equation (i) and (ii) in (iv)
$N = {N_o}{e^ - }^{\left[ {\dfrac{{{{\log }_e}2}}{{{t_1}}} + \dfrac{{{{\log }_e}2}}{{{t_2}}}} \right]t}$
$ \implies {N_o}{e^{ - \left[ {\dfrac{{{t_1} + {t_2}}}{{{t_1}{t_2}}}} \right]\,\log {\,_e}2\,\,t}}$ ……………. (v)
$ \implies N = {N_o}{e^{ - \lambda t}}$ …………… (vi)
Compare equation (v) and (vi)
$\lambda = \dfrac{{{{\log }_e}2}}{{\dfrac{{{t_1}{t_2}}}{{{t_1} + {t_2}}}}} = \dfrac{{{{\log }_e}\,2}}{{{t_3}}}$
Hence ${t_3}$ is effective half life
$ \therefore {t_3} = \dfrac{{{t_1}{t_2}}}{{{t_1} + {t_2}}}$

So, the correct answer is “Option C”.

Note:
Radioactive decay reduces the number of radioactive nuclei over time. In one half-life, the number decreases to half of its original value. Half of what remains decay in the next half-life, and half of those in the next, and so on. This is an exponential decay process and spontaneous.