MHT CET · Physics · Atomic Physics
What is the moment of inertia of the electron moving in second Bohr orbit of hydrogen atom?
[ \(\mathrm{h}=\) Planck's constant, \(\mathrm{m}=\) mass of electron, \(\varepsilon_0=\) permittivity of free space, \(\mathrm{e}=\) charge on electron]
- A \(\frac{4 \varepsilon_0^2 h^4}{\pi^2 m e^4}\)
- B \(\frac{8 m \varepsilon_0^2 h^4}{\pi^2 e^4}\)
- C \(\frac{16 \varepsilon_0^2 h^4}{\pi^2 m e^4}\)
- D \(\frac{\varepsilon_0^2 h^4}{16 \pi^2 m e^4}\)
Answer & Solution
Correct Answer
(C) \(\frac{16 \varepsilon_0^2 h^4}{\pi^2 m e^4}\)
Step-by-step Solution
Detailed explanation
Moment of Inertia \(\mathrm{I}=\mathrm{MR}^2\)
Radius of the \(\mathrm{n}^{\text {th }}\) Bohr orbit is,
\(r_n=\frac{\varepsilon_0 h^2 n^2}{\pi \mathrm{me}^2}\)
For \(\mathrm{n}=2\),
\(r_2=\frac{4 \varepsilon_0 h^2}{\pi m e^2}\)
\(\therefore \quad\) Moment of inertia of the electron in the \(2^{\text {nd }}\) orbit is
\(\begin{aligned}
\text { M.I } & =\mathrm{m} \times\left[\frac{4 \varepsilon_0 \mathrm{~h}^2}{\pi \mathrm{me}^2}\right]^2 \\
& =\mathrm{m} \times \frac{16 \varepsilon_0^2 \mathrm{~h}^4}{\pi^2 \mathrm{~m}^2 \mathrm{e}^4} \\
& =\frac{16 \varepsilon_0^2 \mathrm{~h}^4}{\pi^2 \mathrm{me}^4}
\end{aligned}\)
Radius of the \(\mathrm{n}^{\text {th }}\) Bohr orbit is,
\(r_n=\frac{\varepsilon_0 h^2 n^2}{\pi \mathrm{me}^2}\)
For \(\mathrm{n}=2\),
\(r_2=\frac{4 \varepsilon_0 h^2}{\pi m e^2}\)
\(\therefore \quad\) Moment of inertia of the electron in the \(2^{\text {nd }}\) orbit is
\(\begin{aligned}
\text { M.I } & =\mathrm{m} \times\left[\frac{4 \varepsilon_0 \mathrm{~h}^2}{\pi \mathrm{me}^2}\right]^2 \\
& =\mathrm{m} \times \frac{16 \varepsilon_0^2 \mathrm{~h}^4}{\pi^2 \mathrm{~m}^2 \mathrm{e}^4} \\
& =\frac{16 \varepsilon_0^2 \mathrm{~h}^4}{\pi^2 \mathrm{me}^4}
\end{aligned}\)
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