Question

The isotopic composition of rubidium is ${}^{85}\text{Rb}-$72% and ${}^{87}\text{Rb}-$28%. ${}^{87}\text{Rb}$ is weakly radioactive and decay by $\beta$-emission with a decay constant of $1.1\times {10}^{-11}$ per year. A sample of the mineral pollucite was found to contain 450 mg Rb and 0.72 mg of ${}^{87}\text{Sr}$. The age of mineral pollucite is ____. (Assume whatever is necessary)

• A. $[6.18\times {10}^{8}year]$
• B. $[7.18\times {10}^{8}year]$
• C. $[5.18\times {10}^{8}year]$
• D. $[1.18\times {10}^{9}year]$

Answer 1

Answer: (C)

$[5.18\times {10}^{8}year]$

From given data, the isotopic composition is ${}^{85}Rb=72$%, ${}^{87}Rb=28$%
The mineral sample contained 450 mg of Rb
$\therefore$ Amount of ${}^{87}Rb$ (radioactive)$=450\times \frac{28}{100}=126$ mg of ${}^{87}Rb$
$0.72$ mg of ${}^{87}Sr$ have been formed from $0.72$ mg of ${}^{87}Rb$ by $\beta -$ emission.
$\therefore$ Amount of ${}^{87}Rb$ initially present $=126mg+0.72mg=126.72mg$
Decay constant, $\lambda =1.1\times {10}^{-11}peryear$
Use the relation, $\lambda =\frac{2.303}{t}\log \frac{N_0}{N_t}$
where $t$ is the age of mineral pollucite
$\therefore t=\frac{2.303}{t}\log \frac{126.72}{126}=\frac{2.303}{1.1\times {10}^{-11}}\log \left(1.0057\right)=5.18\times {10}^{8}years$.
Hence, the age of the mineral is $5.18\times {10}^{8}years$.