A) \[0\]
B) \[\frac{e}{\sqrt{{{\varepsilon }_{0}}{{a}_{0}}m}}\]
C) \[\frac{e}{\sqrt{4\pi {{\varepsilon }_{0}}{{a}_{0}}m}}\]
D) \[\frac{\sqrt{4\pi {{\varepsilon }_{0}}{{a}_{0}}m}}{e}\]
Correct Answer: C
Solution :
Centripetal force == force of attraction of nucleus on electron \[\frac{m{{v}^{2}}}{{{a}_{0}}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{e}^{2}}}{a_{0}^{2}}\] \[v=\frac{e}{\sqrt{4\pi {{\varepsilon }_{0}}m{{a}_{0}}}}\]You need to login to perform this action.
You will be redirected in
3 sec