Chemical Kinetics - Result Question 76
75. An experiment requires minimum beta activity produced at the rate of $346$ beta particles per minute. The half-life period of ${ } _{42} Mo^{99}$, which is a beta emitter, is $66.6$ $ h$. Find the minimum amount of ${ } _{42} Mo^{99}$ required to carry out the experiment in $6.909 $ $h$.
$(1989,5 M)$
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Answer:
Correct Answer: 75. $(3.56 \times 10^{-16} $ $g)$
Solution:
- The minimum rate of decay required after $6.909 $ $h$ is $346$ particles $\min ^{-1}$.
$ \begin{aligned} & \Rightarrow \quad \text { Rate }=k N \\ & \Rightarrow \quad N=\frac{\text { Rate }}{k}=\frac{346 \times 66.6 \times 60}{0.693}=1.995 \times 10^{6} \text { atoms } \\ & \Rightarrow \quad k t=\ln \frac{N _0}{N} \Rightarrow \frac{\ln 2}{66.6} \times 6.909=\ln \frac{N _0}{N}=0.0715 \\ & \Rightarrow \quad \frac{N _0}{N}=1.074 \\ & \Rightarrow \quad N _0=1.074 \times N=1.074 \times 1.995 \times 10^{6} \\ & = \quad 2.14 \times 10^{6} \text { atoms of Mo } \\ & \Rightarrow \text { Mass of Mo required }=\frac{2.14 \times 10^{6}}{6.023 \times 10^{23}} \times 99=3.56 \times 10^{-16} g \end{aligned} $
BETA
