Question

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

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

Answer 1

The correct option is A $1080$

$\frac{-d{N}_A}{dt}=\frac{d{N}_B}{dt}=\frac{d{N}_C}{dt}$

${\lambda }_{eq}{N}_A={\lambda }_1{N}_A+{\lambda }_2{N}_A$

${\lambda }_{eq}{N}_A={\lambda }_1{N}_A+{\lambda }_2{N}_A$

${\lambda }_{eq}={\lambda }_1+{\lambda }_2$

Since $\lambda \alpha \frac{1}{t}$

$\therefore$ Half life is related as.

$\frac{1}{{\left(T_{1/2}\right)}_{effective}}=\frac{1}{{\left(T_{1/2}\right)}_{A\to B}}+\frac{1}{{\left(T_{1/2}\right)}_{A\to C}}$

Effective half life = $\frac{1620\times 810}{2430}=540$ yrs.

$\Rightarrow$ It will 2 effective half lives for A to become $\frac{1}{4}th$ of its initial value.

$\therefore$ Time = 1080 years.