Question 389: Atoms & Nuclei

Question

The frequency of revolution of an electron in the $\mathrm{n}^{	ext {th }}$ orbit of hydrogen atom is

Accepted Answer

According to Bohr's atomic model, the radius of the electron's orbit in the $n$th level is:
$$ r_n = \frac{\varepsilon_0 n^2 h^2}{\pi m_e Z e^2} $$
The speed of the electron in the $n$th orbit is:
$$ v_n = \frac{Z e^2}{2 \varepsilon_0 n h} $$
Finding angular frequency:
The angular frequency ($\omega$) is the speed divided by the radius:
$$ \omega = \frac{v}{r} = \frac{\left(\frac{Z e^2}{2 \varepsilon_0 n h}ight)}{\left(\frac{\varepsilon_0 n^2 h^2}{\pi m_e Z e^2}ight)} = \frac{\pi m_e e^4}{2 \varepsilon_0^2 h^3 n^3} $$
Finding frequency of revolution:
The frequency ($f$) is $\omega$ divided by $2\pi$:
$$ f = \frac{\omega}{2\pi} = \frac{\left(\frac{\pi m_e e^4}{2 \varepsilon_0^2 h^3 n^3}ight)}{2\pi} = \frac{m_e e^4}{4 \varepsilon_0^2 h^3 n^3} $$
Relation to $n$:
$$ 	herefore \quad f \propto \frac{1}{n^3} $$

Answer

directly proportional to $n^2$

Answer

inversely proportional to $\mathrm{n}^2$

Answer

directly proportional to $n^3$

Answer

inversely proportional to $\mathrm{n}^3$

Atoms & Nuclei

The frequency of revolution of an electron in the \mathrm{n}^{\text {th }} orbit of hydrogen atom is

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