A) \[\frac{e}{\sqrt{{{\varepsilon }_{0}}{{a}_{0}}m}}\]
B) zero
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 :
Coulomb attraction between electron and proton \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{e}^{2}}}{{{r}^{2}}}\] Centripetal force \[=\frac{m{{v}^{2}}}{r}\] \[\therefore \] \[\frac{m{{v}^{2}}}{r}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{e}^{2}}}{{{r}^{2}}}\] \[\Rightarrow \] \[{{v}^{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{e}^{2}}}{mr}\] For ground state of H-atom \[r={{a}_{0}}\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,v=\frac{e}{\sqrt{4\pi {{\varepsilon }_{0}}m{{a}_{0}}}}\]
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