The magnetic dipole moment associated with an atom due to orbital motion of an electron in fourth stationary orbit of the hydrogen atom is:

$3.708 \times 10^{-23} A m^2$

The correct answer is Option (2) → $3.708 \times 10^{-23} A m^2$

For an electron in the n-th orbit,

$I=\frac{e}{T}=\frac{e}{\frac{2\pi r_n}{v_n}}=\frac{ev_n}{\frac{2\pi r_n}{v_n}}$

Area of circular orbit, $A=\pi{r_n}^2$

$∴μ_n=IA=\frac{ev_n}{2\pi r_n}×\pi{r_n}^2=\frac{eh}{4\pi m}$

$μ_n=\frac{(1.6×10^{-19})(6.63×10^{-34})}{4×(9.11×10^{-31})×(3.14)}$

$=3.708×10^{-23}Am^2$