Question
The orbital speed of electron orbiting around a nucleus in a circular orbit of radius $$50\,pm$$ is $$2.2 \times {10^6}\,m{s^{ - 1}}.$$ Then the magnetic dipole moment of an electron is
A.
$$1.6 \times {10^{ - 19}}A{m^2}$$
B.
$$5.3 \times {10^{ - 21}}A{m^2}$$
C.
$$8.8 \times {10^{ - 25}}A{m^2}$$
D.
$$8.8 \times {10^{ - 26}}A{m^2}$$
Answer :
$$8.8 \times {10^{ - 25}}A{m^2}$$
Solution :
Magnetic dipole moment
$$\eqalign{
& m = iA = \frac{e}{T} \times \pi {r^2} = \frac{e}{{\left( {\frac{{2\pi r}}{v}} \right)}} \times \pi {r^2} = \frac{{erv}}{2}. \cr
& = \frac{{1.6 \times {{10}^{ - 19}} \times 50 \times {{10}^{ - 12}} \times 2.2 \times {{10}^6}}}{2} \cr
& = 8.8 \times {10^{ - 25}}A{m^2}. \cr} $$