Question

Calculate the angular frequency of an electron occupying the second Bohr orbit of ${He}^{+}$ ion.

Answer 1

Velocity of electron $\left(v\right)=\frac{2\pi {Ze}^2}{nh}$ ...(i)
Radius of ${He}^{+}$ ion in an orbit $\left(r_n\right)=\frac{n^2{h}^2}{4{\pi }^{2}m{Ze}^2}$ ...(ii)
Angular frequency or angular velocity $\left(\omega \right)=\frac{v}{r_n}=\frac{2\pi {Ze}^2}{nh}\times \frac{4{\pi }^{2}m{Ze}^2}{n^2{h}^2}=\frac{8{\pi }^{3}m{Z}^2{e}^4}{n^3{h}^3}$
Given, $n=2$, $m=9.1\times {10}^{-28}g$, $Z=2$, $e=4.8\times {10}^{-10}esu$, $h=6.626\times {10}^{-27}erg-sec$
$\therefore \omega =\frac{8\times {\left(\frac{22}{7}\right)}^3\times 2^2\times 9.1\times {10}^{-28}\times {(4.8\times {10}^{-10})}^4}{{\left(2\right)}^3\times {(6.626\times {10}^{-27})}^3}$
$=2.067\times {10}^{16}{sec}^{-1}$.