A) \[\frac{3}{4}{{N}_{0}}\]
B) \[\frac{{{N}_{0}}}{9}\]
C) \[{{N}_{0}}\]
D) None of these
Correct Answer: B
Solution :
As, \[N={{N}_{0}}{{e}^{-\lambda t}}\] \[\frac{{{N}_{0}}}{3}={{N}_{0}}{{e}^{-\,9\lambda }}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{1}{3}={{e}^{-\,9\lambda }}\] Now, after further 9 y \[{N}'\,\,=\,\,{{N}_{0}}{{e}^{-\,18\lambda }}\] \[{N}'\,\,=\,\,{{N}_{0}}{{\left( {{e}^{-\,9\lambda }} \right)}^{2}}\] \[{N}'\,\,=\,\,{{N}_{0}}{{\left( \frac{1}{3} \right)}^{2}}\,\,=\,\,\frac{{{N}_{0}}}{9}\]
You need to login to perform this action.
You will be redirected in
3 sec
