A) 1080
B) 2430
C) 3240
D) 4860
Correct Answer: A
Solution :
\[\frac{-dN}{dt}={{\lambda }_{1}}N+{{\lambda }_{2}}N\] \[\Rightarrow {{\log }_{e}}\frac{N}{{{N}_{0}}}=-({{\lambda }_{1}}+{{\lambda }_{2}})t\] when\[{{N}_{0}}\]is initial number of atoms Here\[{{\lambda }_{1}}=\frac{0.693}{1620}\] and \[{{\lambda }_{2}}=\frac{0.693}{810};\] \[\frac{N}{{{N}_{0}}}=\frac{1}{4}\Rightarrow {{\log }_{e}}\frac{1}{4}=-\left( \frac{0.693}{1620}+\frac{0.693}{810} \right)t\] \[\Rightarrow 2.303[-2\times (.3010)]=-0.693\left[ \frac{810+1620}{1620\times 810} \right]t\] \[\Rightarrow \frac{2\times 1620\times 810}{2430}=t=1080\] years
You need to login to perform this action.
You will be redirected in
3 sec
