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A sample initially contains only $U -238$ isotope of uranium. With time, some of the $U -238$ radioactively decays into $Pb -206$ while the rest of it remains undisintegrated.
When the age of the sample is $P \times 10^8$ years, the ratio of mass of $Pb -206$ to that of $U -238$ in the sample is found to be $7$ . The value of $P$ is. . . . . .
[Given : Half-life of $U-238$ is $4.5 \times 10^9$ years; $\log _e 2=0.693$ ]
$143$
$145$
$150$
$150$
Solution
${ }_{92} U ^{238} \longrightarrow{ }_{82} Pb ^{206}+8{ }_2 He ^4+6{ }_{-1} \beta^0$
$t =0 \quad N _0=\left(\frac{1}{238}+\frac{7}{206}\right)$ moles $t = t N _{ t }=\frac{1}{238}$ moles $x =\frac{7}{206}$ moles
As per $1^{\text {st }}$ order kinetics :
$\lambda t =\ln \frac{ N _0}{ N _{ t }}$
$\frac{\ln 2}{4.5 \times 10^9}=\frac{1}{ t } \ln \frac{\frac{1}{238}+\frac{7}{206}}{\frac{1}{238}}$
$t =\frac{4.5 \times 10^9}{\ln 2} \ln \frac{1872}{206}$
$t =4.5 \times 10^9 \times \frac{\ln (9.08)}{\ell n 2}=4.5 \times 10^9 \times \frac{2.206}{0.693}=143.3 \times 10^8$
