Question 563: Centre of Mass and Rotational Motion

Question

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]

Accepted Answer

Moment of Inertia $$\mathrm{I}=\mathrm{MR}^2$$
Radius of the $$\mathrm{n}^{	ext {th }}$$ Bohr orbit is,
$$\mathrm{r}_{\mathrm{n}}=\frac{\varepsilon_0 \mathrm{~h}^2 \mathrm{n}^2}{\pi \mathrm{me}^2}$$
For $$\mathrm{n}=2$$,
$$\mathrm{r}_2=\frac{4 \varepsilon_0 \mathrm{~h}^2}{\pi \mathrm{me}^2}$$
$$	herefore \quad$$ Moment of inertia of the electron in the $$2^{	ext {nd }}$$ orbit is
$$\begin{aligned}
	ext { M.I } & =m 	imes\left[\frac{4 \varepsilon_0 h^2}{\pi m e^2}ight]^2 \
& =m 	imes \frac{16 \varepsilon_0^2 h^4}{\pi^2 m^2 e^4} \
& =\frac{16 \varepsilon_0^2 h^4}{\pi^2 m e^4}
\end{aligned}$$

Answer

$$\frac{4 \varepsilon_0^2 h^4}{\pi^2 m e^4}$$

Answer

$$\frac{8 m \varepsilon_0^2 h^4}{\pi^2 e^4}$$

Answer

$$\frac{16 \varepsilon_0^2 h^4}{\pi^2 m e^4}$$

Answer

$$\frac{\varepsilon_0^2 h^4}{16 \pi^2 \mathrm{me}^4}$$

Centre of Mass and Rotational Motion

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]

Choose the correct answer: