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Question Bank [ 2022 ] — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsQuestion Bank [ 2022 ]4 Marks
Question
Using the expression for the radius of orbit for the Hydrogen atom, show that the linear speed varies inversely to the principal quantum number $n$ the angular speed varies inversely to the cube of principal quantum number $n.$
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Answer

According to Bohr's second postulate,
$ mr _{ n } v _{ n }=\frac{ nh }{2 \pi}$
$\therefore m ^2 v _{ n }^2 r _{ n }^2=\frac{ n ^2 h ^2}{4 \pi^2}$
$\therefore v _{ n }^2=\frac{ n ^2 h ^2}{4 \pi^2 m ^2 r _{ n }^2} $
Substituting, $r_n=\frac{\varepsilon_0 h^2 n^2}{\pi m Z e^2}$ in above relation,
$ v _{ n }^2=\frac{ n ^2 h ^2}{4 \pi^2 m ^2} \times\left(\frac{\pi mZe }{\varepsilon_0 h ^2 n ^2}\right)^2$
$=\frac{ n ^2 h ^2}{4 \pi^2 m ^2} \times \frac{\pi^2 m ^2 Z ^2 e ^4}{\varepsilon_0^2 h ^4 n ^4}$
$=\frac{ Z ^2 e ^4}{4 \varepsilon_0^2 h ^2 n ^2} $
$ \therefore v _{ n }^2 \propto \frac{1}{ n ^2}$
$\Rightarrow v _{ n } \propto \frac{1}{ n } $
Expression for angular speed:
Since, $v_n=r_n \omega$ and $r_n=\frac{\varepsilon_0 h^2 n^2}{\pi m_e e^2}$
$ \therefore \omega=\frac{ v _{ n }}{ r _{ n }}=\left(\frac{ e ^2}{2 \varepsilon_0 h }\right) \frac{1}{ n } / \frac{\varepsilon_0 h ^2 n ^2}{\pi m _{ e } e ^2}$
$\therefore \omega=\frac{ e ^2}{2 \varepsilon 0 hn } \times \frac{\pi m _{ e } e ^2}{\varepsilon_0 h ^2 n ^2}=\left(\frac{\pi m _{ e } e ^4}{2 \varepsilon_0^2 h ^3}\right) \frac{1}{ n ^3}$
$\Rightarrow \omega \propto \frac{1}{ n ^3} $

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