Question

${}^{87}\text{Rb}$ undergoes $\beta$-decay $(t_{1/2}=4.7\times {10}^{10}year)$ to form ${}^{87}\text{Sr}$. An analysis of the rock sample shows relative number of ${}^{87}\text{Sr}$ and ${}^{87}\text{Rb}$ atoms $0.052:1$. If all the ${}^{87}\text{Sr}$ is formed from $\beta$-decay of ${}^{87}\text{Rb}$, The age of the rock is :

• A. $[2.21\times {10}^{9}year]$
• B. $[3.43\times {10}^{9}year]$
• C. $[4.13\times {10}^{9}year]$
• D. none of these

Answer 1

Answer: (B)

$[3.43\times {10}^{9}year]$

The beta decay follows first-order kinetics.

Rate constant, $k=$ $\frac{0.693}{t_{1/2}}$

Given, $t_{1/2}=4.7\times {10}^{10}$ years

$k=\frac{0.693}{4.7\times {10}^{10}\text{years}}$

$k=1.47\times {10}^{-11}$

Also, $t=\frac{2.303}{k}\times log\frac{N_o}{N_t}$

$N_o=1.052$

$N_t=1.0$

$t=\frac{2.303}{1.47\times {10}^{-11}}\times log\frac{1.052}{1.0}$

Hence, the age of the rock, $t=3.43\times {10}^9$ years.