Question

An electron in an atom revolved around the nulceus in an orbit of radius $0.53\stackrel{o}{A}$. If the frequency of revolution of an electron is $9\times {10}^9$ MHz, calculate the orbital angular momentum. [Given: Charge on an electron $=1.6\times {10}^{-19}$C; Gyromagnetic ratio $=8.8\times {10}^{10}C/kg;\pi =3.142$].

Answer 1

Given : $r=0.53$ $A^o$ and $\nu =9\times {10}^{15}$ Hz

Gyromagnetic ratio $\gamma =8.8\times {10}^{10}C/kg$

Magnetic moment $\mu =e\nu \left(\pi r^2\right)$

$\therefore$ $\mu =(1.6\times {10}^{-19})\times (9\times {10}^{15})\times \pi \times (0.53\times {10}^{-10}{)}^2$

$⟹$ $\mu =12.7\times {10}^{-24}$ $A{m}^2$

Orbital angular momentum $L=\frac{\mu }{\gamma }$

$\therefore$ $L=\frac{12.7\times {10}^{-24}}{8.8\times {10}^{10}}=1.44\times {10}^{-34}$ $kg{m}^3/s$