Question

The electron in the hydrogen atom is moving at a speed of $1.1 \times {10^6}{\text{m}}{{\text{s}}^{ - 1}}$ in an orbit of radius $2.12{{\text{A}}^o}$ . Calculate the magnetic moment of the revolving electron.

Answer 1

Hint:We are given the circular orbit of radius and the velocity of the electron revolving around the hydrogen atom. We will use the formula for the magnetic dipole moment of a revolving electron and substitute the given values.

Complete step by step answer:
The magnetic moment is the magnetic strength of the current in circular loops or permanent magnets. Given $r = 2.12{{\text{A}}^o}$ the radius of the electron and $v = 1.1 \times {10^6}{\text{m}}{{\text{s}}^{ - 1}}$ which is the velocity of the revolving electron.The electronic charge of an electron is constant which is equal to
$e = 1.6 \times {10^{ - 19}}$

Now we need to find the current produced due to the revolving electron which is given by
$i = ef$
where $f$ is the frequency of the electron revolving and is expressed as
$f = \dfrac{\omega }{{2\pi }}$
where $\omega $ is the angular frequency.
And we know the angular frequency or the angular velocity is linear velocity divided by the radius of the circular orbit, hence we have
$f = \dfrac{v}{{2\pi r}}$ using $v = r\omega $
Substituting the expression for frequency in the current expression we get
$i = \dfrac{{ev}}{{2\pi r}}$

Now the dipole magnetic moment of a revolving electron is given by
$\mu = iA$
Where $A$ is the cross-sectional area of the flowing current.
Substituting for the current and cross-sectional area we get
$\mu = \dfrac{{ev}}{{2\pi r}} \times \pi {r^2} = \dfrac{{evr}}{2}$
Substituting the values given to us in the question we get
$\mu = \dfrac{{1.6 \times {{10}^{ - 19}} \times 1.1 \times {{10}^6} \times 2.12 \times {{10}^{ - 10}}}}{2} \\
\therefore \mu = 1.86 \times {10^{ - 23}}{\text{Am}}$

Therefore the magnetic moment of the electron is $1.86 \times {10^{ - 23}}{\text{ Am}}$.

Note:We derived the whole expression for magnetic moments in terms of the physical quantities whose values were given to us in the question. We calculated the cross-sectional area of the electron based on the shape of the electron. The magnetic moment varies with respect to the shape of the revolving body or the charged body which is producing the magnetic dipole moment.