Q165 / 180 · KCET 2020 · Physics

The period of revolution of an electron revolving in nth orbit of H-atom is proportional to

  • A
    $n^{2}$
  • B
    $\frac{1}{n}$
  • Correct answer
    C
    $n^{3}$
  • D
    independent of $n$

Solution

The time period of revolution of an electron revolving in $n$th orbit of $\mathrm{H}$-atom is given by $\begin{array}{ll} & T_{n}=\frac{2 \pi r_{n}}{v_{n}}=\left(\frac{4 \varepsilon_{0}^{2} h^{2}}{m e^{4}}\right) \frac{n^{3}}{z^{2}} \\ \Rightarrow \quad & T_{n} \propto n^{3} \end{array}$

Hence, the correct answer is option: 'C'.

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