Question

The isotopes ${U^{238}}$ and ${U^{235}}$ occur in nature in the ratio $140:1$. Assuming that at the time of earth formation, they were present in equal ratio, the estimated age of earth will be:
(The half-life period of ${U^{238}}$ and ${U^{235}}$ are $4.5 \times {10^9}$ and $7.13 \times {10^8}$ years, respectively.)
A. $6.04 \times {10^9}$ years
B. $6.25 \times {10^9}$ years
C. $6.59 \times {10^9}$ years
D. none of these

Answer 1

The decomposition of uranium follows first order kinetics. To answer this question, you should recall the formula for decrease in the concentration of reactants in a first order reaction with time.
Formula used:
${C_t} = {C_0}{e^{ - kt}}$ …..(1.1)
For first order reaction, $k = \dfrac{{\ln 2}}{T}$ …..(1.2)
Where, ${C_t}$is concentration of reactant at time $t$, ${C_0}$ is initial concentration of reactant, $t$ is time, $k$ is rate constant and $T$ is half-life of the reactant.

Complete step by step answer:
We are given the question that initial concentrations of ${U^{238}}$ and ${U^{235}}$ are equal.
Let the initial concentrations of ${U^{238}}$ and ${U^{235}}$ be ${C_0}$ and final concentrations of ${U^{238}}$ and ${U^{235}}$ at time $t$ be ${C_1}$ and ${C_2}$ respectively.
We know that ${C_t} = {C_0}{e^{ - kt}}$.
Hence,
$\Rightarrow {C_1} = {C_0}{e^{ - {k_1}t}}$ …..(1.3)
$\Rightarrow {C_2} = {C_0}{e^{ - {k_2}t}}$ …..(1.4)
where ${k_1}$ and ${k_2}$ are the rate constants for the decomposition of ${U^{238}}$ and ${U^{235}}$ respectively.
We know that, $\dfrac{{{C_1}}}{{{C_2}}} = \dfrac{{140}}{1}$ …..(1.5)
Dividing equation (1.3) by equation (1.4), we get,
$\dfrac{{{C_1}}}{{{C_2}}} = \dfrac{{{C_0}{e^{ - {k_1}t}}}}{{{C_0}{e^{ - {k_2}t}}}} = \dfrac{{{e^{ - {k_1}t}}}}{{{e^{ - {k_2}t}}}} \\\ \Rightarrow \dfrac{{{C_1}}}{{{C_2}}} = {e^{{k_2}t - {k_1}t}} \\\$
Taking natural log on both sides of the equation, we get
$\left( {{k_2} - {k_1}} \right)t = \ln \left( {\dfrac{{{C_1}}}{{{C_2}}}} \right)$
Rearranging to get $t$,
$t = \dfrac{{\ln \left( {\dfrac{{{C_1}}}{{{C_2}}}} \right)}}{{\left( {{k_2} - {k_1}} \right)}}$ ……(1.6)
From equation (1.2), we know that,
${k_1} = \dfrac{{\ln 2}}{{{T_1}}}$ and ${k_2} = \dfrac{{\ln 2}}{{{T_2}}}$, where ${T_1} = 4.5 \times {10^9}years$and ${T_2} = 7.13 \times {10^8}years$ are half-lives of ${U^{238}}$and ${U^{235}}$respectively.
Substituting these values in equation (1.6), We get,
$\Rightarrow t = \dfrac{{\log \left( {140} \right)}}{{\log 2}}\left[ {\dfrac{{{T_1} \times {T_2}}}{{{T_1} - {T_2}}}} \right]$ $t = \dfrac{{\ln \left( {\dfrac{{{C_1}}}{{{C_2}}}} \right)}}{{\ln 2\left( {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right)}} = \dfrac{{\log \left( {140} \right)}}{{\log 2\left( {\dfrac{{{T_1} - {T_2}}}{{{T_1}.{T_2}}}} \right)}}$
$\Rightarrow t = \dfrac{{\log \left( {140} \right)}}{{\log 2}}\left[ {\dfrac{{{T_1} \times {T_2}}}{{{T_1} - {T_2}}}} \right]$
$\Rightarrow t = \dfrac{{2.1461}}{{0.3010}}\left[ {\dfrac{{4.5 \times {{10}^9} \times 7.13 \times {{10}^8}}}{{4.5 \times {{10}^9} - 7.13 \times {{10}^8}}}} \right]$
$\therefore t = 6.04 \times {10^9}years$

Hence, the correct option is A.

Note:
For a first order reaction, half-life is independent of the initial concentration of the reactant. The concentration of reactant decreases exponentially with time in a first order equation. If we are given the value of ${C_0}$ and ${C_t}$ at different time instants, the value of $k$ can be calculated for different time instants by using the first order law. If the reaction for which the data is given is a first order reaction, then all values of $k$ will be approximately equal to each other.