Modern Physics 3 Question 40
39. At a given instant there are $25 %$ undecayed radioactive nuclei in a sample. After $10 s$ the number of undecayed nuclei reduces to $12.5 %$. Calculate
$(1996,3$ M)
(a) mean life of the nuclei,
(b) the time in which the number of undecayed nuclei will further reduce to $6.25 %$ of the reduced number.
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Solution:
- (a) In $10 s$, number of nuclei has been reduced to half (25% to $12.5 %)$.
Therefore, its half-life is $t _{1 / 2}=10 s$
Relation between half-life and mean life is
$$ \begin{aligned} & t _{\text {mean }}=\frac{t _{1 / 2}}{\ln (2)}=\frac{10}{0.693} s \\ & t _{\text {mean }}=14.43 s \end{aligned} $$
(b) From initial $100 %$ to reduction till $6.25 %$, it takes four half lives.
$$ \begin{array}{ll} 100 % & \stackrel{t _{1 / 2}}{\rightarrow} 50 % \stackrel{t _{1 / 2}}{\rightarrow} 25 % \stackrel{t _{1 / 2}}{\rightarrow} 12.5 % \stackrel{t _{1 / 2}}{\rightarrow} 6.25 % \\ \therefore & t=4 t _{1 / 2}=4(10) s=40 s \\ & t=40 s \end{array} $$
