A) \[\text{1}0\text{8}0\text{yr}~~\]
B) \[\text{234}0\text{yr}\]
C) \[\text{486}0\text{ yr}\]
D) \[~\text{324}0\text{ yr}\]
Correct Answer: A
Solution :
Since, from Rutherford-Soddy law, the number of atoms left after half lives is given by \[N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{n}}\]where, \[{{N}_{0}}\] is the original number of atoms. The number of half lives, \[\text{n =}\frac{\text{time of decay}}{\text{effective half-life}}\] Relation between effective disintegration constant (\[\lambda \]) and half life (T) \[\lambda =\frac{\ln 2}{T}\] \[\therefore \] \[{{\lambda }_{1}}+{{\lambda }_{2}}=\frac{\ln 2}{{{T}_{1}}}+\frac{\ln 2}{{{T}_{2}}}\] Effective half life, \[\frac{1}{T}=\frac{1}{{{T}_{1}}}+\frac{1}{{{T}_{2}}}=\frac{1}{1620}+\frac{1}{810}\] \[\frac{1}{T}=\frac{1+2}{1620}\Rightarrow T=540yr\] \[\therefore \] \[n=\frac{T}{540}\] \[\therefore \] \[N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{t/540}}\Rightarrow \frac{N}{{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{2}}\] \[={{\left( \frac{1}{2} \right)}^{t/540}}\] \[\Rightarrow \frac{t}{540}=2\Rightarrow t=2\times 540=1080yr\]
You need to login to perform this action.
You will be redirected in
3 sec
