• question_answer                 In the Bohr's model of a hydrogen atom, the centripetal force is furnished by the Coulomb attraction between the proton and the electron. If ${{a}_{0}}$ is the radius of the ground state orbit, m is the mass and e is the charge on the electron, ${{\varepsilon }_{0}}$ is the vacuum permittivity, the speed of the electron is:                            A)                 zero       B)                 $\frac{e}{\sqrt{{{\varepsilon }_{0}}{{a}_{0}}m}}$                            C)                 $\frac{e}{\sqrt{4\pi {{\varepsilon }_{0}}{{a}_{0}}m}}$                   D)                 $\frac{\sqrt{4\pi {{\varepsilon }_{0}}{{a}_{0}}m}}{e}$

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                 $\therefore v=\frac{e}{\sqrt{4\pi {{\varepsilon }_{0}}m{{a}_{0}}}}$