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 Newtons second law, a radially inward centripetal force is needed to the electron which is being provided by the Coulombs attraction between the proton and electron. |
Coulombs 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, \[r={{a}_{0}}\] |
\[\therefore \] \[v=\frac{e}{\sqrt{4\pi {{\varepsilon }_{0}}m{{a}_{0}}}}\] |
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