A) \[4\lambda \]
B) \[2\lambda \]
C) \[1/2\lambda \]
D) \[1/4\lambda \]
Correct Answer: C
Solution :
[c] \[{{\lambda }_{A}}=5\lambda \] and \[{{\lambda }_{B}}=\lambda \] At \[t=0,\] s\[{{({{N}_{0}})}_{A}}={{({{N}_{0}})}_{B}}\] Given, \[\frac{{{N}_{A}}}{{{N}_{B}}}={{\left( \frac{1}{e} \right)}^{2}}\] According to radioactive decay, \[\frac{N}{{{N}_{0}}}={{e}^{-\lambda t}}\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{{N}_{A}}}{{{({{N}_{0}})}_{A}}}={{e}^{-\lambda \,\,{{A}^{t}}}}\] and \[\frac{{{N}_{B}}}{{{({{N}_{0}})}_{B}}}={{e}^{-\lambda \,{{B}^{t}}}}\] From (1) and (2), \[\frac{{{N}_{A}}}{{{N}_{B}}}={{e}^{-(5\lambda -\lambda )t}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,{{\left( \frac{1}{e} \right)}^{2}}={{e}^{-4\lambda t}}={{\left( \frac{1}{e} \right)}^{4\lambda t}}\Rightarrow \,4\lambda t=2\,\,\therefore \,t=\frac{1}{2\lambda }.\]
You need to login to perform this action.
You will be redirected in
3 sec
