Nuclear Chemistry - Result Question 24
####24. ${ } _{92} U^{238}$ is radioactive and it emits $\alpha$ and $\beta$ particles to form ${ } _{82} Pb^{206}$. Calculate the number of $\alpha$ and $\beta$ particles emitted in this conversion.
An ore of ${ } _{92} U^{238}$ is found to contain ${ } _{92} U^{238}$ and ${ } _{82} Pb^{206}$ in the weight ratio of $1: 0.1$. The half-life period of ${ } _{92} U^{238}$ is $4.5 \times 10^{9}$ yr. Calculate the age of the ore.
(2000)
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Answer:
Correct Answer: 24. $\left(7.12 \times 10^{8} yr\right)$
Solution:
- ${ } _{92} U^{238} \longrightarrow{ } _{82} Pb^{206}+8{ } _2 He^{4}+6{ } _{-1} e^{0}$
Present : $N _0-N \quad N$
Given, $\frac{w(U)}{w(Pb)}=\frac{1}{0.1}=10$
$\Rightarrow \quad \frac{N(U)}{N(Pb)}=\frac{10}{238} \times \frac{206}{1}=\frac{N _0-N}{N}$
$\Rightarrow \quad \frac{N}{N _0-N}=\frac{238}{2060} \Rightarrow \frac{N _0}{N _0-N}=1+\frac{238}{2060}=\frac{2298}{2060}$
Now, applying first order rate law
$$ \begin{gathered} \left(\frac{\ln 2}{t _{1 / 2}}\right) t=\ln \left(\frac{N _0}{N _0-N}\right) \Rightarrow t=\frac{\left(t _{1 / 2}\right)}{\log 2} \log \left(\frac{N _0}{N _0-N}\right) \ =\frac{4.5 \times 10^{9}}{0.3} \log \frac{2298}{2060}=7.12 \times 10^{8} yr \end{gathered} $$
