Question

n hydrogen atom, an electron is making $6.6\times10^{15}\,rps$ around the nucleus in an orbit of radius 0.523 A. Calculate the equivalent magnetic dipole moment.

• A. $9.06\times10^{-18}\,Am^2$
• B. $9.06\times10^{-24}\,Am^2$
• C. $3.06\times10^{-24}\,Am^2$
• D. $18.12\times10^{-24}\,Am^2$

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 = $(1.6 \times 10^{-19}) \times (6.6 \times 10^{15}) \times \frac{22}{7} \times (0.523 \times 10^{-10} )^2$
= $9.06 \times 10^{-24} \, Am^2$