A) 4860 yr
B) 3240 yr
C) 2340 yr
D) 1080 yr
Correct Answer: D
Solution :
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 \[\frac{1}{T}={{\frac{1}{T}}_{1}}+{{\frac{1}{T}}_{2}}=\frac{1}{1620}+\frac{1}{810}\] \[\frac{1}{T}=\frac{1+2}{1620}\Rightarrow T=540\,yr\] \[\therefore \] \[n=\frac{t}{540}\] \[\therefore \] \[N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{t/540}}\] \[\Rightarrow \] \[\frac{N}{{{N}_{0.}}}={{\left( \frac{1}{2} \right)}^{2}}={{\left( \frac{1}{2} \right)}^{t/540}}\] \[\Rightarrow \] \[\frac{t}{540}=2\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{t=2 }\!\!\times\!\!\text{ 540=1080yr}\]
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