A) 0
B) \[_{7}{{N}^{14}}\]
C) \[\frac{1}{4}\]
D) \[\frac{\sqrt{4\pi {{\varepsilon }_{0}}{{a}_{0}}m}}{e}\]
Correct Answer: C
Solution :
\[\frac{m{{v}^{2}}}{{{a}_{0}}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{e}^{2}}}{a_{0}^{2}}\Rightarrow v=\frac{e}{\sqrt{4\pi {{\varepsilon }_{0}}{{a}_{0}}m}}\]You need to login to perform this action.
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