Question

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

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

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
$n=\frac{timeofdecay}{effectivehalf-life}$
Relation between effective disintegration constant ( $\lambda$ ) and half-life $(T)$ is
$\lambda =\frac{ln2}{T}$
$\therefore {\lambda }_1+{\lambda }_2=\frac{ln2}{T_1}+\frac{ln2}{T_2}$
Effective half -life
$\frac{1}{T}=\frac{1}{T_1}+\frac{1}{T_2}=\frac{1}{1620}+\frac{1}{810}$
$\frac{1}{T}=\frac{1+2}{1620}\Rightarrow T\Rightarrow 540yr$
$\therefore n=\frac{t}{540}$
$\Rightarrow$ $\frac{t}{540}=2$
$\Rightarrow t=2\times 540=1080yr$