In the Bohr model of the hydrogen atom,the centripetal force is provided by the Coulomb attraction between the proton and the electron. If $r_0$ is the radius of the ground state orbit,$m$ is the mass,$e$ is the charge on the electron,and $\varepsilon_0$ is the permittivity of vacuum,the speed of the electron is:

  • A
    zero
  • B
    $\frac{e}{\sqrt{\varepsilon_0 r_0 m}}$
  • C
    $\frac{e}{\sqrt{4 \pi \varepsilon_0 r_0 m}}$
  • D
    $\frac{\sqrt{4 \pi \varepsilon_0 r_0 m}}{e}$

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To calculate the size of a hydrogen ion $(H^-)$ using the Bohr's model,we assume that its two electrons move in an orbit such that they are always on diametrically opposite sides of the nucleus. With each electron having the angular momentum $\hbar = h / 2\pi$ and taking electron interaction into account,the radius of the orbit in terms of the Bohr's radius of hydrogen atom $a_B = \frac{4\pi\varepsilon_0\hbar^2}{me^2}$ is

For an electron moving in the $n^{\text{th}}$ Bohr orbit,the de Broglie wavelength of the electron is:

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Reason $(R)$: Magnetic moment of the revolving electron,$\mu \propto n$.

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