Modern Physics Ques 140
- 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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Answer:
Correct Answer: 140.(a) $14.43$ $s \quad$ (b) $40$ $s$
Solution:
Formula:
- (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} $
BETA
