Question

Assume that the isotope $U^{238}$ and $U^{235}$ are found in the ratio 64 : 1 presently at a planet. At the time of the creation of the planet, they were present in an equal ratio.What is the age of the planet (in million years ${10}^6$ ). The half-life period of $U^{238}$ and $U^{235}$ are $4.45\times {10}^9$ and $7.0\times {10}^8$ year respectively.

Answer 1

Let the age of the planet be = t years
Given,
$U_o^{238}:U_o^{235}=1:1$
$U_t^{238}:U_t^{235}=64:1$
Now, ${\lambda }^{238}.t=ln\frac{U_0^{238}}{U_t^{238}}$ and ${\lambda }^{235}.t=ln\frac{U_0^{235}}{U_t^{235}}$

Hence,$({\lambda }^{238}-{\lambda }^{235})t=ln(\frac{U_0^{238}}{U_t^{238}}\times \frac{U_t^{235}}{U_0^{235}})$
$\Rightarrow ln2\times (\frac{1}{4.45\times {10}^9}-\frac{1}{7\times {10}^8})=ln\left(\frac{1}{64}\right)$

$\Rightarrow ln2\times \left(\frac{4.45-0.7}{4.45\times 0.7\times {10}^9}\right)t=ln\left(\frac{64}{1}\right)\Rightarrow ln2\times \left(\frac{3.75}{4.45\times 0.7\times {10}^9}\right)t=6ln2$

$\Rightarrow t=\frac{4.45\times 0.7\times {10}^9\times 6}{3.75}=4984\times {10}^6$ years.