Question 675: Atoms & Nuclei

Question

For a radioactive material, its activity A and rate of change of its activity R are defined as $$A =  - {{dN} \over {dt}}$$ and $$R =  - {{dA} \over {dt}}$$, where N(t) is the number of nuclei at time t. Two radioactive source P(mean life $$	au $$) and Q (mean life 2$$	au $$) have the same activity at t = 0. Their rate of change of activities at t = 2$$	au $$ are R
P
and R
Q
, respectively. If $${{{R_P}} \over {{R_Q}}} = {n \over e}$$, then the value of n is

Accepted Answer

Law of radioactivity : $$N = {N_0}{e^{ - \lambda t}}$$ where $$\lambda$$ = decay constant
Activity $$\left| A ight| = \left| {{{ - dN} \over {dt}}} ight| = {N_0}\lambda {e^{ - \lambda t}}$$
Rate of activity $$R = {{d|A|} \over {dt}} = {N_0}{\lambda ^2}{e^{ - \lambda t}}$$
At t = 0, A
1
= A
2
. Therefore,
$${N_{OP}}{\lambda _P} = {N_{OQ}}{\lambda _Q}$$
At $$t = 2	au ,\,{{{R_P}} \over {{R_Q}}} = {\left( {{{{\lambda _P}} \over {{\lambda _Q}}}} ight)^2}\left( {{{{N_{OP}}} \over {{N_{OQ}}}}} ight){{{e^{ - \lambda P(25)}}} \over {{e^{ - \lambda Q(25)}}}} = {{{\lambda _P}} \over {{\lambda _Q}}}{e^{({\lambda _Q} - {\lambda _P})25}}$$
Since mean life is given by $$	au  = {1 \over \lambda }$$.
Therefore,
$${{{R_P}} \over {{R_Q}}} = {{{\lambda _P}} \over {{\lambda _Q}}}{e^{{{\left( {{1 \over {25}} - {1 \over 5}} ight)}^{25}}}} = {{{\lambda _P}} \over {{\lambda _Q}}}{e^{ - 1}}$$
$${{{R_P}} \over {{R_Q}}} = {{{\lambda _\{ }} \over {{\lambda _Q}}}{1 \over e} = {n \over e}$$
$$n = {{{\lambda _P}} \over {{\lambda _Q}}} = {{2	au } \over 	au } = 2$$

Atoms & Nuclei

Integer / Numeric Answer Type