Question 17: Atoms & Nuclei

Question

In a radioactive material the activity at time t
1
, is A
1
and at a later time t
2
, it is A
2
. If the decay constant of the material is $$\lambda$$, then

Accepted Answer

From radioactive decay law,
$$ - {{dN} \over {dt}} \propto N$$ or $$ - {{dN} \over {dt}} = \lambda N$$
Thus, $$A =  - {{dN} \over {dt}}$$ or $$A = \lambda N$$
$$ \Rightarrow A = \lambda {N_0}{e^{ - \lambda t}}$$ ...... (i)
Where, $${A_0} = \lambda {N_0}$$ is the activity of the radioactive material at time, t = 0
At time, $${A_1} = {A_0}\,{e^{ - \lambda {t_1}}}$$ ..... (ii)
At time, $${A_2} = {A_0}\,{e^{ - \lambda {t_2}}}$$ ..... (iii)
On dividing Eq. (ii) by Eq. (iii), we have
$${{{A_1}} \over {{A_2}}} = {{{e^{ - \lambda {t_1}}}} \over {{e^{ - \lambda {t_2}}}}} = {e^{ - \lambda ({t_1} - {t_2})}}$$
$$ \Rightarrow {A_1} = {A_2}\,{e^{ - \lambda ({t_1} - {t_2})}}$$

Answer

$${A_1} = {A_2}\,{e^{ - \lambda ({t_1} - {t_2})}}$$

Answer

$${A_1} = {A_2}\,{e^{\lambda ({t_1} - {t_2})}}$$

Answer

$${A_1} = {A_2}({t_2}/{t_1})$$

Answer

$${A_1} = {A_2}$$

Atoms & Nuclei

In a radioactive material the activity at time t 1 , is A 1 and at a later time t 2 , it is A 2 . If the decay constant of the material is $\lambda$, then

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