Question

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

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

Answer 1

Answer: (D)

1080 yr

Answer 2

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 $n=\frac{\text{time}\,\text{of}\,\text{decay}}{\text{effective}\,\text{half}-\text{life}}$
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 $\frac{1}{T}={{\frac{1}{T}}_{1}}+{{\frac{1}{T}}_{2}}=\frac{1}{1620}+\frac{1}{810}$
$\frac{1}{T}=\frac{1+2}{1620}\Rightarrow T=540\,yr$
$\therefore$ $n=\frac{t}{540}$
$\therefore$ $N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{t/540}}$
$\Rightarrow$ $\frac{N}{{{N}_{0.}}}={{\left( \frac{1}{2} \right)}^{2}}={{\left( \frac{1}{2} \right)}^{t/540}}$
$\Rightarrow$ $\frac{t}{540}=2$
$\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{t=2 }\\!\\!\times\\!\\!\text{ 540=1080yr}$