Question

In the hydrogen atom the electron moves around the proton with a speed of $2.0\times {10}^6{ms}^{-1}$ in a circular orbit of radius $5.0\times {10}^{-11}$ What is the equivalent dipole moment ?

• A. $2\times {10}^{-24}{Am}^2$
• B. $4\times {10}^{-24}{Am}^2$
• C. $8\times {10}^{-24}{Am}^2$
• D. $16\times {10}^{-24}{Am}^2$

Answer 1

The correct option is C $8\times {10}^{-24}{Am}^2$

$\begin{array}{c}v=2\times {10}^6\\ t=\frac{2\pi r}{v}=\frac{2\pi \times 5\times {10}^{-11}}{2\times {10}^{-6}}\\ =5\pi \times {10}^{-17}\\ i=\frac{q}{t}=\frac{1.6\times {10}^{-19}}{5\pi \times {10}^{-17}}\\ =\frac{1.6\times {10}^{-2}}{5\pi }\\ =\frac{16\times {10}^{-3}}{5\pi }\end{array}$

$\begin{array}{c}\mu =iA\\ =\frac{16\times {10}^{-3}}{5\pi }\times \pi 25\times {10}^{-22}\\ =80\times {10}^{-25}\\ =8\times {10}^{-24}\end{array}$

Option $C$ is correct .