Question

n hydrogen atom, an electron is making $6.6\times 10^{15}rps$ around the nucleus in an orbit of radius 0.523 A. Calculate the equivalent magnetic dipole moment.

• A. $9.06\times 10^{-18}A{m}^2$
• B. $9.06\times 10^{-24}A{m}^2$
• C. $3.06\times 10^{-24}A{m}^2$
• D. $18.12\times 10^{-24}A{m}^2$

Answer 1

Answer: (B)

Dipole moment of an atom M = I $\times$ A = $\frac{e}{T}\pi r^2$
or M = $e\times \nu \times \pi r^2$
Here $e$ is the charge on electron, is revolutions per second, $r$ is the radius.
or M = $(1.6\times 10^{-19})\times (6.6\times 10^{15})\times \frac{22}{7}\times (0.523\times 10^{-10}{)}^2$
= $9.06\times 10^{-24}A{m}^2$