Question
If the electron revolving around the nucleus in a radius $$'r'$$ with orbital speed $$'v'$$ has magnetic moment $$\frac{{evr}}{2}.$$ Hence, using Bohr's postulate of the quantization of angular momentum obtain the magnetic moment $$\left( M \right)$$ of hydrogen atom in its ground state and current $$\left( I \right)$$ due to revolution of electron.
A.
$$M = \frac{{eh}}{{4\pi m}},I = \frac{{eV}}{{2\pi r}}$$
B.
$$M = \frac{{2eh}}{{5\pi m}},I = \frac{{eV}}{{4\pi r}}$$
C.
$$M = \frac{h}{{\pi m}},I = \frac{e}{{\pi r}}$$
D.
$$M = \frac{{eh}}{{\pi m}},I = \frac{{eV}}{{\pi r}}$$
Answer :
$$M = \frac{{eh}}{{4\pi m}},I = \frac{{eV}}{{2\pi r}}$$
Solution :
$$\eqalign{
& I = \frac{e}{T} = \frac{{eV}}{{2\pi r}} \cr
& {\text{so,}}\,M = \frac{{eV}}{{2\pi r}} \times \pi {r^2} = \frac{{evr}}{2} \cr} $$
According to Bohr’s theory angular momentum
$$\eqalign{
& mvr = \frac{{nh}}{{2\pi }}\,\,{\text{or}}\,\,vr = \frac{{nh}}{{2\pi m}} \cr
& {\text{so,}}\,\,M = \frac{{neh}}{{4\pi m}} \cr} $$
For the ground state $$n = 1\,{\text{so,}}\,\,M = \frac{{eh}}{{4\pi m}}$$