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Surface Energy

The weight ba...

Question

The weight based ratio of $U^{238}$ and $P b^{226}$ in a sample of rock is 4 : 3. If the half life of $U^{238}$ is $4.5 \times 10^{9}$ years, then the age of rock is:

9.0 \times 10^{9}

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6.3 \times 10^{9} y e a r s

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4.5 \times 10^{9} y e a r s

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3.78 \times 10^{9} y e a r s

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Solution

The correct option is C $3.78 \times 10^{9} y e a r s$
$t = \frac{l n (\frac{A_{0}}{A_{t}})}{k} = \frac{l n (\frac{\frac{4}{238} + \frac{3}{226}}{\frac{4}{238}})}{l n 2} t_{\frac{1}{2}}$
$k = \frac{l n_{2}}{t_{\frac{1}{2}}} ≃ 3 \cdot 78 \times 10^{9} y e a r s$

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Similar questions

Q. The weight based ratio of

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and

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in a sample of rock in

4 : 3

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The weight based ratio of U238 and Pb226 in a sample of rock is 4 : 3. If the half life of U238 is 4.5×109 years, then the age of rock is:

The correct option is C 3.78×109yearst=ln(A0At)k=ln(4238+32264238)ln2t12k=ln2t12≃3⋅78×109years

The weight based ratio of $U^{238}$ and $P b^{226}$ in a sample of rock is 4 : 3. If the half life of $U^{238}$ is $4.5 \times 10^{9}$ years, then the age of rock is:

The correct option is C $3.78 \times 10^{9} y e a r s$
$t = \frac{l n (\frac{A_{0}}{A_{t}})}{k} = \frac{l n (\frac{\frac{4}{238} + \frac{3}{226}}{\frac{4}{238}})}{l n 2} t_{\frac{1}{2}}$
$k = \frac{l n_{2}}{t_{\frac{1}{2}}} ≃ 3 \cdot 78 \times 10^{9} y e a r s$