Question

A radioactive material by simultaneous emission of two particles with respective half-lives $1620$ and $810$ year. The time, in years, after which one-fourth of the martial remains is:

• A. $1080$
• B. $2430$
• C. $3240$
• D. $4860$

Answer 1

The correct option is A $1080$

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 original number of atoms.
The number of half-life $n=\frac{timeofdecay}{efectivehalf-life}$

Relation between effective disintegration constant$\left(\lambda \right)$ 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 $\frac{1}{T}=\frac{1}{T_1}+{\frac{1}{T}}_2=\frac{1}{1620}+\frac{1}{810}$

$\frac{1}{T}=\frac{1+2}{1620}$

$T=540yr$

Therefore, $n=\frac{t}{540}$

$N=N_0{\left(\frac{1}{2}\right)}^{\frac{t}{540}}$

$\frac{N}{N_0}={\left(\frac{1}{2}\right)}^2={\left(\frac{1}{2}\right)}^{\frac{t}{540}}$

$\Rightarrow \frac{t}{540}=2$

$\Rightarrow t=2\times 540=1080yr$