Question
A radioactive sample consists of two distinct species having equal number of atoms initially. The mean life time of one species is $$\tau $$ and that of the other is $$5\tau .$$ The decay products in both cases are stable. A plot is made of the total number of radioactive nuclei as a function of time. Which of the following figures best represent the form of this plot?
A.
B.
C.
D.
Answer :
Solution :
$$\eqalign{ & {N_1} = {N_0}{e^{ - {\lambda _1}t}} = {N_0}{e^{ - \frac{t}{\tau }}}\,......\left( {\text{i}} \right) \cr & {\text{as}}\,\tau = \frac{1}{{{\lambda _1}}} \cr & {N_2} = {N_0}{e^{ - {\lambda _2}t}} = {N_0}{e^{ - \frac{t}{{5\tau }}}}\,......\left( {\text{ii}} \right)\,{\text{as }}5\tau = \frac{1}{{{\lambda _2}}} \cr} $$
Adding (i) and (ii) we get
$$N = {N_1} + {N_2} = {N_0}\left( {{e^{ - \frac{t}{\tau }}} + {e^{ - \frac{t}{{5\tau }}}}} \right)$$
(A) is NOT the correct option as there is a time $$\tau $$ for which $$N$$ is constant which means for time $$\tau $$ there is no process of radioactivity which does not makes sense. (B) and (C) shows intermediate increase in the number of radioactive atom which is IMPOSSIBLE as $$N$$ will only decrease exponentially.
$$\eqalign{ & {N_1} = {N_0}{e^{ - {\lambda _1}t}} = {N_0}{e^{ - \frac{t}{\tau }}}\,......\left( {\text{i}} \right) \cr & {\text{as}}\,\tau = \frac{1}{{{\lambda _1}}} \cr & {N_2} = {N_0}{e^{ - {\lambda _2}t}} = {N_0}{e^{ - \frac{t}{{5\tau }}}}\,......\left( {\text{ii}} \right)\,{\text{as }}5\tau = \frac{1}{{{\lambda _2}}} \cr} $$
Adding (i) and (ii) we get
$$N = {N_1} + {N_2} = {N_0}\left( {{e^{ - \frac{t}{\tau }}} + {e^{ - \frac{t}{{5\tau }}}}} \right)$$
(A) is NOT the correct option as there is a time $$\tau $$ for which $$N$$ is constant which means for time $$\tau $$ there is no process of radioactivity which does not makes sense. (B) and (C) shows intermediate increase in the number of radioactive atom which is IMPOSSIBLE as $$N$$ will only decrease exponentially.