A sample of uranium rock from moon contains equimolar amounts of $U^{238}$ and $P b^{206}$ . The age of the rock will be $(t_{\frac{1}{2}} f o r U^{238} = 4.5 \times 10^{9} y r s)$

$4.5 \times 10^{9} y r s .$

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$9.0 \times 10^{9} y r s .$

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$13.5 \times 10^{9} y r s .$

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$2.25 \times 10^{9} y r s .$

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Solution

The correct option is A
$4.5 \times 10^{9} y r s .$

$1 + \frac{P b}{U} = 1 + 1 = 2 = \frac{U + P b}{U} = \frac{N_{0}}{N}$
Hence, $t = \frac{2.303}{λ} l o g \frac{N_{0}}{N}; λ = \frac{0.693}{t_{\frac{1}{2}}} = \frac{0.693}{4.5 \times 10^{9}}$
Thus, $t = \frac{2.303 \times 4.5 \times 10^{9}}{0.693} l o g 2 = 4.5 \times 10^{9} y r .$

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A sample of uranium rock from moon contains equimolar amounts of $U^{238}$ and $P b^{206}$ . The age of the rock will be $(t_{\frac{1}{2}} f o r U^{238} = 4.5 \times 10^{9} y r s)$

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A sample of uranium rock from moon contains equimolar amounts of U238 and Pb206. The age of the rock will be (t12for U238=4.5×109yrs)

The correct option is A 4.5×109yrs. 1+PbU=1+1=2=U+PbU=N0N Hence, t=2.303λlogN0N;λ=0.693t12=0.6934.5×109 Thus, t=2.303×4.5×1090.693log 2=4.5×109yr.

A sample of uranium rock from moon contains equimolar amounts of $U^{238}$ and $P b^{206}$ . The age of the rock will be $(t_{\frac{1}{2}} f o r U^{238} = 4.5 \times 10^{9} y r s)$