Question

A radioactive material decays by simultaneous emission of two particles with half-lives $1620 \, yr$ and $810 \, yr$ respectively. The time (in $yr$ ) after which one-fourth of the material remains, is

• A. $4860yr$
• B. $3240yr$
• C. $2340yr$
• D. $1080 \, yr$

Answer 1

Answer: (D)

From Rutherford-Soddy law, the number of atoms left after $n$ half-lives is given by
$ \, \, \, N=N_{0}\left(\frac{1}{2}\right)^{n}$
Where, $N_{0}$ is the original number of atoms.
The number of half-life
$ \, n=\frac{t i m e \, o f \, d e c a y}{e f f e c t i v e \, h a l f - l i f e}$
Relation between effective disintegration constant ( $\lambda $ ) and half-life $(T)$ is
$ \, \, \, \lambda =\frac{ln 2}{T}$
$\therefore \, \, \lambda _{1}+\lambda _{2}=\frac{ln 2}{T_{1}}+\frac{ln ⁡ 2}{T_{2}}$
Effective half -life
$ \, \, \frac{1}{T}=\frac{1}{T_{1}}+\frac{1}{T_{2}}=\frac{1}{1620}+\frac{1}{810}$
$ \, \frac{1}{T}=\frac{1 + 2}{1620}\Rightarrow T\Rightarrow 540 \, yr$
$\therefore \, \, n= \, \frac{t}{540}$
$\Rightarrow $ $\frac{t}{540}=2$
$\Rightarrow \, \, \, t= \, 2 \, \times \, 540 \, = \, 1080 \, yr$