The time period of revolution of electron in its ground state orbit in a hydrogen atom
Question:

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:

  1. (1) $1.6 \times 10^{14}$

  2. (2) $7.8 \times 10^{14}$

  3. (3) $6.2 \times 10^{15}$

  4. (4) $5.6 \times 10^{12}$


Correct Option: 2,

Solution:

(2) For first excited state $n^{\prime}=3$

Time period $T \propto \frac{n^{3}}{z^{2}}$

$\Rightarrow \frac{T_{2}}{T_{1}}=\frac{n^{\prime 3}}{n^{3}}$

$\therefore T_{2}=8 T_{1}=8 \times 1.6 \times 10^{-16} S$

$\therefore$ Frequency, $v=\frac{1}{T_{2}}=\frac{1}{8 \times 1.6 \times 10^{-16}}$

$\approx 7.8 \times 10^{14} \mathrm{~Hz}$

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