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 $5.0 \times 10^{-11} m$. what is 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

Answer: (C)

we know that the magnetic moment of a current loop is $M=\mid A . $
Equivalent current due to revolution of electron is
$I=e / T=e / 2 k r / v=e x v / 2 k r$.
but $A=k r^{2}$.
Therefore $M=(e \times v / 2 k r) \times k r^{2}$
$=1.6 \times 10^{-19} \times 5 \times 10^{-}{ }^{11} \times 2 \times 10^{6} / 2$
$=8 \times 10^{-24} Am ^{-2}$