Question

The isotopes ${}^{238}U$ and ${}^{235}U$ occur in nature in the weight ration $140:1$. It is assumed that initially they were found in equal weight. If half life $\left(t_{1/2}\right)$ of ${}^{238}U=4.5\times {10}^9$ and $t_{1/2}$ $235U=5\times {10}^5$ year respectively then the age of earth is $(log7=0.846;log2=0.3)$

• A. $4.02\times {10}^{9}year$
• B. $2.01\times {10}^{9}year$
• C. $8.72\times {10}^{9}year$
• D. $0.35\times {10}^{7}year$

Answer 1

The correct option is D $0.35\times {10}^{7}year$

$\frac{\text{mass of}{U}^{238}}{\text{mass of}{U}^{235}}=\frac{140}{1}$

$\begin{array}{c}\text{Initially they have equal weight}\\ T_{1/2}\text{(half life)}{U}^{238}=4.5\times {10}^9\text{year}\\ T_{1/2}\text{for}{U}^{235}=5\times {10}^5\text{year}\end{array}$

$\begin{array}{c}\text{after time t, Let us say}{U}^{238}\text{left}=N_1\\ \text{and}{U}^{235}\text{left}=N_2\end{array}$

$\text{Now,}$

$N_1=N_0{e}^{-{\lambda }_{1}t}$

$\text{and}{N}_2=N_0{e}^{-{\lambda }_{2}t}$

$\begin{array}{c}\because \frac{N_1}{N_2}=e^{({\lambda }_{2}t-{\lambda }_{1}t)}\\ \text{and}\lambda =\frac{\ln 2}{T_{1/2}}\end{array}$

$\text{Now,}\frac{N_1}{N_2}=e^{t({\lambda }_2-{\lambda }_1)}$

$\begin{array}{c}t=\frac{\ln \left(N_1/N_2\right)}{{\lambda }_2-{\lambda }_1}\\ t=\frac{\ln \left(N_1/N_2\right)}{\ln 2(\frac{1}{T_2}-\frac{I}{T_1})}\\ t=\frac{\ln \left(N_1/N_2\right)\times T_1\times T_2}{\ln 2(T_1-T_2)}\end{array}$

$\text{Putting all values ,}$

$\begin{array}{cc}t& =\frac{\ln \left(140\right)\times 4.5\times {10}^9\times 5\times {10}^5}{\ln 2(4.5\times {10}^9-5\times {10}^5)}\\ t& =0.35\times {10}^7\text{year}\end{array}$