A) zero
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 :
Key Idea: According to the Newton?s second law, a radially inward centripetal force is needed to the electron which is being provided by the Coulomb?s attraction between the proton and electron. Coulomb?s attraction between the positive proton and negative electron \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{e}^{2}}}{{{r}^{2}}}\] Centripetal force has magnitude \[F=\frac{m{{v}^{2}}}{r}\] As per key idea, \[\frac{m{{v}^{2}}}{r}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{e}^{2}}}{{{r}^{2}}}\] \[\Rightarrow {{v}^{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{e}^{2}}}{mr}\] \[\Rightarrow \] \[v=\frac{e}{\sqrt{4\pi {{\varepsilon }_{0}}mr}}\] For ground state of H-atom, r = a0 \[\therefore v=\frac{e}{\sqrt{4\pi {{\varepsilon }_{0}}m{{a}_{0}}}}\]
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