Using Bohr's model, the orbital period of electron in hydrogen atom in $n^{\text {th }}$ orbit is…

Using Bohr's model, the orbital period of electron in hydrogen atom in $n^{\text {th }}$ orbit is $\left(\varepsilon_0=\right.$ permittivity of vacuum, $h=$ Planck's constant, $m=$ mass of electron, $e=$ electronic charge)
  1. $\frac{4 \varepsilon_0^2 n^3 h^3}{m e^4}$
  2. $\frac{4 \varepsilon_0^2 n^2 h^3}{m e^3}$
  3. $\frac{4 \varepsilon_0 n h^3}{m e^2}$
  4. $\frac{4 \varepsilon_0 n^2 h^2}{m e^2}$

Solution

The orbital period of revolution of electron in $n^{\text {th }}$ orbit is $T_n=\frac{2 \pi r_n}{v_n}$ Bohr radius for $n^{\text {th }}$ orbit $r_n=\left(\frac{h^2 \varepsilon_0}{\pi m e^2}\right) \frac{n^2}{Z}$ and $v_n=\left(\frac{e^2}{2 h \varepsilon_0}\right) \frac{Z}{n}$. $\therefore T_n=2 \pi \frac{h^2 \varepsilon_0 n^2}{\pi m e^2 Z} \times \frac{2 h \varepsilon_0 n}{e^2 Z}=\frac{4 \varepsilon_0^2 n^3 h^3}{m e^4 Z^2}$ For hydrogen atom, $Z=1$ $\therefore T_n==\frac{4 \varepsilon_0^2 n^3 h^3}{m e^4}$

Asked in: MHT CET 2022 (10 Aug Shift 2)

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