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

The Correct Option is 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 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} $