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Question: A radioactive sample decays in two modes. In one mode its half life is \({t_1}\) and in the other mo...

A radioactive sample decays in two modes. In one mode its half life is t1{t_1} and in the other mode its half life is t2{t_2}. Find the overall half life
(A) t1+t2{t_1} + {t_2}
(B) t1+t22\dfrac{{{t_1} + {t_2}}}{2}
(C) t1t2t1+t2\dfrac{{{t_1}{t_2}}}{{{t_1} + {t_2}}}
(D) t1t2t1t2\dfrac{{{t_1}{t_2}}}{{{t_1} - {t_2}}}

Explanation

Solution

Half life is defined as the number of atoms reduced to half of its present value.
t12=loge2λ{t_{^{\dfrac{1}{2}}}} = \dfrac{{\log {\,_e}\,2}}{\lambda }

Complete step by step answer:
λ1=loge2t1{\lambda _1} = \dfrac{{{{\log }_e}2}}{{{t_1}}} …………. (i)
λ2=loge2t2{\lambda _2} = \dfrac{{{{\log }_e}2}}{{{t_2}}} ………… (ii)
Where t1{t_1} and t2{t_2} are their respective half life and λ1{\lambda _1} and λ2{\lambda _2} are decay constants.
So the decay rate of quantity NN is given by
dNdt=Nλ1+Nλ2- \dfrac{{dN}}{{dt}} = N{\lambda _1} + N{\lambda _2}
    dNN=(λ1+λ2)dt\implies \dfrac{{dN}}{N} = - ({\lambda _1} + {\lambda _2})dt …………. (iii)
Integrate equation (iii)
NoNdNN=ot(λ1+λ2)dt\int_{{N_o}}^N {\dfrac{{dN}}{N}} = - \int_o^t {({\lambda _1} + {\lambda _2})\,dt}
    logeNNoN=(λ1+λ2)[t]ot\implies \left| {{{\log }_e}N} \right|_{{N_o}}^N = - ({\lambda _1} + {\lambda _2})[t]_o^t
    logeNlogeNo=(λ1+λ2)t\implies {\log _e}N - {\log _e}{N_o} = - ({\lambda _1} + {\lambda _2})t
    logeNNo=(λ1+λ2)t\implies {\log _e}\dfrac{N}{{{N_o}}} = - \left( {{\lambda _1} + {\lambda _2}} \right)t
Take antilog on both sided
NNo=e(λ1+λ2)t\dfrac{N}{{{N_o}}} = {e^{ - ({\lambda _1} + {\lambda _2})t}}
    N=Noe(λ1+λ2)t\implies N = {N_o}{e^{ - ({\lambda _1} + {\lambda _2})t}} …………. (iv)
Use equation (i) and (ii) in (iv)
N=Noe[loge2t1+loge2t2]tN = {N_o}{e^ - }^{\left[ {\dfrac{{{{\log }_e}2}}{{{t_1}}} + \dfrac{{{{\log }_e}2}}{{{t_2}}}} \right]t}
    Noe[t1+t2t1t2]loge2t\implies {N_o}{e^{ - \left[ {\dfrac{{{t_1} + {t_2}}}{{{t_1}{t_2}}}} \right]\,\log {\,_e}2\,\,t}} ……………. (v)
    N=Noeλt\implies N = {N_o}{e^{ - \lambda t}} …………… (vi)
Compare equation (v) and (vi)
λ=loge2t1t2t1+t2=loge2t3\lambda = \dfrac{{{{\log }_e}2}}{{\dfrac{{{t_1}{t_2}}}{{{t_1} + {t_2}}}}} = \dfrac{{{{\log }_e}\,2}}{{{t_3}}}
Hence t3{t_3} is effective half life
t3=t1t2t1+t2\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.