An electron in the ground state of hydrogen atom in revolving in anticlockwise direction in a circular orbit of radius R (see fig)
If the expression for the orbital magnetic moment of the electron is $M = \frac{e h}{X π m}$ . Find X?

Question

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Answer 1

In ground state $(n = 1)$ according to Bohr's theory:
$m v R = \frac{h}{2 π}$ or $v = \frac{h}{2 π m R}$

Now time period, $T = \frac{2 π R}{v} = \frac{2 π R}{h / 2 π m R} = \frac{4 π ^{2} m R ^{2}}{h}$

Magnetic moment, $M = i A$
Where, $I = \frac{c h a r g e}{time period} = \frac{e}{\frac{4 π ^{2} m R ^{2}}{h}} = \frac{e h}{4 π ^{2} m R ^{2}}$
and $A = π R^{2}$
$∴ M = (π R^{2}) (\frac{e h}{4 π ^{2} m R ^{2}})$ or $M = \frac{e h}{4 π m}$
Direction of magnetic moment $M$ is perpendicular to the plane of orbit.

Answer 2

In ground state

(n = 1)

according to Bohr's theory:

m v R = \frac{h}{2 π}

or

v = \frac{h}{2 π m R}

Now time period,

T = \frac{2 π R}{v} = \frac{2 π R}{h / 2 π m R} = \frac{4 π ^{2} m R ^{2}}{h}

Magnetic moment,

M = i A

Where,

I = \frac{c h a r g e}{time period} = \frac{e}{\frac{4 π ^{2} m R ^{2}}{h}} = \frac{e h}{4 π ^{2} m R ^{2}}

and

A = π R^{2}

∴ M = (π R^{2}) (\frac{e h}{4 π ^{2} m R ^{2}})

or

M = \frac{e h}{4 π m}

Direction of magnetic moment

M

is perpendicular to the plane of orbit.