The ratio of the magnetic dipole moment to the angular momentum of the electron in the 1^st orbit of hydrogen atom is

CONCEPT:

• Magnetic dipole moment. It is a magnetic property of electric current loops or magnets.
• The quantity of magnetic dipole moment is equal to the amount of current flowing in the loop multiplied by the area that the loop encompasses.

Magnetic dipole moment (μ) = IA

where I is current and A is the area.

• Angular momentum: the quantity of a rotating body, which is the product of its moment of inertia and its angular velocity.

Angular momentum (L) = m v r

where m is the mass of the body, v is velocity and r is the distance from the rotating point.

From Bohr's Model of Atom:A

​angular momentum (L) = mvr = nh/2π

v = nh/2πmr

Time period T = 2πr/v

current I = q/T = e/T = e/(2πr/v) = \(enh \over 4\pi^2mr^2\)

where n is the orbit number, h is plank constant, m is the mass of electron, r is the radius of orbit, v is the velocity of the electron, e is the charge on one electron.

CALCULATION:

Magnetic dipole moment (μ) = IA

μ = \({enh \over 4\pi^2mr^2} \times A\) = \({enh \over 4\pi^2mr^2} \times \pi r^2\)

angular momentum (L) = nh/2π

\(\frac{μ}{L}=\frac{e}{2m}\)

So the correct answer is option 1.