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:
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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