A) 9 : 8
B) 1 : 8
C) 8 : 1
D) 3 : 8
Correct Answer: A
Solution :
\[{{N}_{A}}={{N}_{OA}}{{e}^{-\lambda t}}=\frac{{{N}_{OA}}}{{{2}^{t/{{t}_{1/2}}}}}=\frac{{{N}_{OA}}}{{{2}^{6}}}\] \[\therefore \]Number of nuclei decayed \[={{N}_{OA}}-\frac{{{N}_{OA}}}{{{2}^{6}}}=\frac{63{{N}_{OA}}}{64}\] \[{{N}_{B}}={{N}_{OB}}{{e}^{-\lambda t}}=\frac{{{N}_{OB}}}{{{2}^{t/{{t}_{1/2}}}}}=\frac{{{N}_{OB}}}{{{2}^{3}}}\] \[\therefore \] Number of nuclei decayed \[{{N}_{OB}}-\frac{{{N}_{OB}}}{{{2}^{3}}}=\frac{7{{N}_{OB}}}{8}\] Since \[{{N}_{OA}}={{N}_{OB}}\] \[\therefore \]Ratio of decayed numbers of nuclei \[A\And B=\frac{63{{N}_{OA}}\times 8}{64\times 7{{N}_{OB}}}=\frac{9}{8}\]
You need to login to perform this action.
You will be redirected in
3 sec
