Question

In an old rock, the mass ratio of ${}_{92}^{238}U$ to ${}_{82}^{206}Pb$ is found to be $595:103$. The age of the rock is: (Mean life of ${}_{92}^{238}U$ is $T_0$)

• A. $T_0\ln 1.2$
• B. $T_0\ln \frac{698}{595}$
• C. $T_0\frac{\ln 1.2}{\ln 2}$
• D. $T_0\frac{\ln \frac{698}{595}}{\ln 2}$

Answer 1

The correct option is B $T_0\ln \frac{698}{595}$

We know
$D=P(e^{\lambda T}-1)\left(\frac{P}{D}\right)(e^{\lambda T}-1)=1\frac{P}{D}{e}^{\lambda T}=(\frac{P}{D}+1)\ln \left(\frac{P}{D}\right)+\lambda T=\ln \left(\frac{698}{595}\right)$
similarly
$\frac{1}{\lambda }\ln (\frac{D}{P}+1)=t$
so, $\overline{T}.\ln (\frac{D}{P}+1)=t$
$t=T_o\ln \left(\frac{698}{595}\right)⟶$ Option B.