Question

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

• A. $6.04 \times 10^{9}$years
• B. $5.69 \times 10^{10}$ years
• C. $6.69 \times 10^{11}$years
• D. $6.69 \times 10^{9}$ years

Answer 1

Answer: (A)

$6.04 \times 10^{9}$years

Answer 2

Let the age of the earth be t years

For $238U$ $\lambda_{1} \times t = 2.303\log\frac{N_{0}U_{238}}{NU_{238}}$ ......(i)

For $235U$ $\lambda_{2} \times t = 2.303\log\frac{N_{0}U_{235}}{NU_{235}}$ ......(ii) Subtracting eq. (ii) from eq. (i)

$t(\lambda_{1} - \lambda_{2}) = 2.303\left\lbrack \log\frac{N_{0}U_{238}}{NU_{238}} - \log\frac{N_{0}U_{235}}{NU_{235}} \right\rbrack = 2.303\log\frac{N_{0}U_{238}}{NU_{235}} \cdot \frac{N_{0}U_{235}}{NU_{238}}t\left( \frac{0.693}{4.5 \times 10^{9}} - \frac{0.693}{7.13 \times 10^{8}} \right) = 2.303\log\frac{1}{140}$

$= - (2.303)(2.1461)$

$= - 4.9425$

$t = 6.04 \times 10^{9}$ years