Modern Physics 3 Question 34
35. A rock is $1.5 \times 10^{9} yr$ old. The rock contains ${ }^{238} U$ which disintegrates to form ${ }^{206} Pb$. Assume that there was no ${ }^{206} Pb$ in the rock initially and it is the only stable product formed by the decay. Calculate the ratio of number of nuclei of ${ }^{238} U$ to that of ${ }^{206} Pb$ in the rock. Half-life of ${ }^{238} U$ is $4.5 \times 10^{9} yr$. $\left(2^{1 / 3}=1.259\right)$
(2004, 2M)
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Solution:
- Let $N _0$ be the initial number of nuclei of ${ }^{238} U$.
After time $t, N _U=N _0 \quad \frac{1}{2}{ }^{n}$
Here $n$ = number of half-lives $=\frac{t}{t _{1 / 2}}=\frac{1.5 \times 10^{9}}{4.5 \times 10^{9}}=\frac{1}{3}$
$N _U=N _0 \frac{1}{2}^{\frac{1}{3}}$ and $N _{Pb}=N _0-N _U=N _0 \quad 1-\frac{1}{2}^{1 / 3}$
$$ \therefore \quad \frac{N _U}{N _{Pb}}=\frac{\frac{1}{2}^{1 / 3}}{1-\frac{1}{2}^{3}}=3.861 $$
