Question

In a chemical analysis of a rock, the mass ratio of two radioactive isotopes is found to be $100:1$. The mean lives of the two isotopes are $4\times {10}^9\text{years}$ and $2\times {10}^9\text{years}$, respectively. If it is assumed that at the time of formation, atoms of both the isotopes were in equal proportion, calculate the age of rock. Ratio of the atomic weight of the two isotopes is $1.02:1$.
Given ${\log }_e\left(100/1.02\right)=4.58$

• A. $1.83\times {10}^9\text{years}$
• B. $1.82\times {10}^8\text{years}$
• C. $1.83\times {10}^{10}\text{years}$
• D. $1.82\times {10}^{11}\text{years}$

Answer 1

The correct option is C $1.83\times {10}^{10}\text{years}$

Given:
$\frac{m_1}{m_2}=\frac{100}{1};\frac{M_1}{M_2}=\frac{1.02}{1}$
$\overline{T_1}=4\times {10}^9\text{years};\overline{T_2}=2\times {10}^9\text{years}$

At $t=0,\frac{(N_1{)}_0}{(N_2{)}_0}=\frac{1}{1}$

At any time, $t$,

$\frac{N_1}{N_2}=\frac{m_1/M_1}{m_2/M_2}=\frac{m_1}{m_2}\times \frac{M_2}{M_1}=\frac{100}{1.02}$

From law of radioactivity,

$N=N_0{e}^{-\lambda t}=N_0{e}^{-t\times \frac{1}{\overline{T}}}$

$\therefore \frac{N_1}{N_2}=(\frac{N_1}{N_2}{)}_0{e}^{(\frac{t}{\overline{T_2}}-\frac{t}{\overline{T_1}})}$

Substituting the value

$\frac{100}{1.02}=1\times e^{(\frac{t}{2\times {10}^9}-\frac{t}{4\times {10}^9})}$

Taking ${\log }_e$ both side, we get

${\log }_e\frac{100}{1.02}=\frac{t}{4\times {10}^9}$

$\therefore t=4.58\times 4\times {10}^9=1.83\times {10}^{10}\text{years}$

Hence, option $\left(\text{C}\right)$ is the correct answer.