- Home
- Standard 12
- Physics
13.Nuclei
hard
The activity of a sample of a radioactive material is ${A_1}$ at time ${t_1}$ and ${A_2}$ at time ${t_2}$ $({t_2} > {t_1}).$ If its mean life $T$, then
A
${A_1}{t_1} = {A_2}{t_2}$
B
${A_1} - {A_2} = {t_2} - {t_1}$
C
${A_2} = {A_1}{e^{({t_1} - {t_2})/T}}$
D
${A_2} = {A_1}{e^{({t_1}/{t_2})T}}$
Solution
(c) $A = {A_0}{e^{ – \lambda t}} = {A_0}{e^{ – t/\tau }};$ where $\tau = $ mean life
So $ \Rightarrow \Delta L = \frac{h}{{2\pi }}({n_2} – {n_1})$
==>${A_0} = \frac{{{A_1}}}{{{e^{ – {t_1}/T}}}} = {A_1}{e^{{t_1}/T}}$
$\therefore {A_2} = {A_0}{e^{ – t/T}} = ({A_1}{e^{{t_1}/T}})\,{e^{ – {t_2}/T}}$
$ \Rightarrow {A_2} = {A_1}{e^{({t_1} – {t_2})/T}}$
Standard 12
Physics
