SATHEE: Modern Physics 3 Question 34

Modern Physics 3 Question 34

35. A rock is $1.5 \times 10^{9} y r$ old. The rock contains $^{238} U$ which disintegrates to form $^{206} P b$ . Assume that there was no $^{206} P b$ 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 $^{206} P b$ in the rock. Half-life of $^{238} U$ is $4.5 \times 10^{9} y r$ . $(2^{1 / 3} = 1.259)$

(2004, 2M)

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Solution:

Let $N_{0}$ be the initial number of nuclei of $^{238} U$ .

After time $t, N_{U} = N_{0} \frac{1}{2}^{n}$

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 $N_{P b} = N_{0} - N_{U} = N_{0} 1 - {\frac{1}{2}}^{1 / 3}$

$∴ \frac{N_{U}}{N_{P b}} = \frac{{\frac{1}{2}}^{1 / 3}}{1 - {\frac{1}{2}}^{3}} = 3.861$

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Modern Physics 3 Question 34

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