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There are $0.618 μ g$ of $^{206} P b$ and $0.238 μ g$ of $^{238} U$ in a rock. If $T_{50}$ of $^{238} U$ is $1.5 \times 10^{9} y r$ , age of the rock is
$1.5 \times 10^{9} y r$
$1.5 \times 10^{7} y r$
$5 \times 10^{9} y r$
$3 \times 10^{9} y r$

1.5 \times 10^{7} y r

5 \times 10^{9} y r

3 \times 10^{9} y r

1.5 \times 10^{9} y r

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Solution

Verified by Toppr

As given,
$[P b] = \frac{0.618 \times 10^{- 6}}{206} m o l = 3 \times 10^{- 9} m o l$
$[U] = \frac{0.238 \times 106 - 6}{238} m o l = 1 \times 10^{- 9} m o l$
For rock-dating, $(1 + \frac{[P b]}{[U]}) = (2)^{y}$
$∴ y = 2 = \frac{t}{T_{50}}$
$∴ t = 3.0 \times 10^{9} y e a r s$

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There are $0.618 μ g$ of $^{206} P b$ and $0.238 μ g$ of $^{238} U$ in a rock. If $T_{50}$ of $^{238} U$ is $1.5 \times 10^{9} y r$ , age of the rock is
$1.5 \times 10^{9} y r$
$1.5 \times 10^{7} y r$
$5 \times 10^{9} y r$
$3 \times 10^{9} y r$

As given,
$[P b] = \frac{0.618 \times 10^{- 6}}{206} m o l = 3 \times 10^{- 9} m o l$
$[U] = \frac{0.238 \times 106 - 6}{238} m o l = 1 \times 10^{- 9} m o l$
For rock-dating, $(1 + \frac{[P b]}{[U]}) = (2)^{y}$
$∴ y = 2 = \frac{t}{T_{50}}$
$∴ t = 3.0 \times 10^{9} y e a r s$