Question

The isotopes $U^{238}$ 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, the estimated age of the age of earth will be :
(The half-life period of $U^{238}$ and $U^{235}$ are $4.5\times {10}^9$ and $7.13\times {10}^8$ years, respectively.)

• A. $6.04\times {10}^9$ years
• B. $6.25\times {10}^9$ years
• C. $6.59\times {10}^9$ years
• D. non of these

Answer 1

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

In nature, $\frac{N_{U^{238}}}{N_{U^{235}}}=\frac{140}{1}$ (at t = t)
At the time of earth formation.
$\frac{N_{0_{U^{238}}}}{N_{0_{U^{235}}}}\times \frac{N_{U^{235}}}{N_{U^{238}}}=\frac{1}{140}$

$\therefore$ For $U^{238},\frac{N_{0_{U^{238}}}}{N_{0_{U^{235}}}}=e^{{\lambda }^{238t}}$ .... (1)

$\therefore$ For $U^{235},\frac{N_{0_{U^{238}}}}{N_{U^{235}}}=e^{{\lambda }^{235t}}$ .... (2)

$\frac{N_{0_{U^{238}}}}{N_{0_{U^{235}}}}\times \frac{N_{U^{235}}}{N_{U^{238}}}=e^{({\lambda }^{238}-{\lambda }^{235})t}$

$\frac{1}{140}=e^{({\lambda }^{238}-{\lambda }^{235})t}$

$e({\lambda }^{238}-{\lambda }^{235}{)}^t=lo{g}_{e}1-lo{g}_{e}140$

$\left[\frac{0.693}{4.5\times {10}^9}-\frac{0.693}{7.13\times {10}^8}\right]t=-2.303log140$
$=-4.9416$
$t=6.04\times {10}^9$ years