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JEE Mains · Physics · STD 12 - Atoms
  • JEE Mains 2020
  • Hard
  • MCQ

The time period of revolution of electron in its ground state orbit in a hydrogen atom is \(1.6 \times 10^{-16} \;\mathrm{s}\). The frequency of revolution of the electron in its first excited state (in \(s^{-1}\) ) is:

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Answer & Solution

Correct Answer

(C) \(7.8 \times 10^{14}\)

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Detailed explanation

Time period of revolution of electron in \(\mathrm{n}^{\text {th }}\) orbit \(\mathrm{T}=\frac{2 \pi \mathrm{r}}{\mathrm{V}}=\frac{2 \pi \mathrm{a}_{0}\left(\frac{\mathrm{n}^{2}}{\mathrm{Z}}\right)}{\mathrm{V}_{0}\left(\frac{\mathrm{Z}}{\mathrm{n}}\right)}\)

\(\Rightarrow \mathrm{T} \propto \frac{\mathrm{n}^{3}}{\mathrm{Z}^{2}}\)

\(\frac{\mathrm{T}_{2}}{\mathrm{T}_{1}}=\frac{(2)^{3}}{(1)^{3}}=8 \Rightarrow \mathrm{T}_{2}=8 \times 1.6 \times 10^{-16}\)

Now frequency \(f_{2}=\frac{1}{T_{2}}=\frac{10^{16}}{8 \times 1.6} \approx 7.8 \times 10^{14} \mathrm{Hz}\)

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