Question

The isotope $U^{238}$ and $U^{235}$ occur in nature in the ratio 140: 1. Assuming that at the time of earth formation, they were present in equal ratio, an estimation of the age of earth is.
The half-life period of $U^{238}$ and $U^{235}$ are $4.5\times {10}^9$ and $7.13\times {10}^8$ year respectively.

• A. $6.04\times {10}^9$
• B. $8.74\times {10}^9$
• C. $7.34\times {10}^9$
• D. None of these

Answer 1

The correct option is A $6.04\times {10}^9$

In nature, $\frac{N_{U^{238}}}{N_{U^{235}}}=\frac{140}{1}att=t$......(1)
At the time of earth formation,
$\frac{N_{0U^{238}}}{N_{0U^{235}}}=\frac{1}{1}att=0$......(2)
Divide equation (2) by equation (1)
$\therefore \frac{N_{0U^{238}}}{N_{0U^{235}}}\times \frac{N_{U^{235}}}{N_{U^{238}}}=\frac{1}{140}$......(3)
$\therefore For{U}^{238}:\frac{N_{0U^{238}}}{N_{U^{238}}}=e^{{\lambda }^{238}.t}$ ... (4)
$For{U}^{235}:\frac{N_{0U^{235}}}{N_{U^{235}}}=e^{{\lambda }^{235}.t}$ ... (5)
Divide equation (4) by equation (5)
$\therefore \frac{N_{0U^{238}}}{N_{0U^{235}}}\times \frac{N_{U^{235}}}{N_{U^{238}}}=e^{\left({\lambda }^{238}-{\lambda }^{235}\right)t}$
Substitute equation (3) in the above equation $\frac{1}{140}=e^{\left({\lambda }^{238}-{\lambda }^{235}\right)t}$
$({\lambda }^{238}-{\lambda }^{235})t={\log }_{e}1-{\log }_{e}140$
But $\lambda =\frac{0.693}{t_{1/2}}$
Hence, $[\frac{0.693}{4.5\times {10}^9}-\frac{0.693}{7.13\times {10}^8}]t=-2.303{\log }_{10}140$
$=-4.9416$
Hence, the age of the earth is $t=6.04\times {10}^9$