Question

Show that the electron revolving around the nucleus in a radius $r$ with orbital speed $v$ has magnetic moment $evr/2$. Hence, using Bohr's postulate of the quantization of angular momentum, obtain the expression for the magnetic moment of a hydrogen atom in its ground state.

Answer 1

Hint:In this problem, to determine the magnetic moment of the electron revolving around the nucleus, apply the formula of magnetic moment $M = IA$ as the orbit is equivalent to the magnetic shell of the magnetic moment. Hence, use this calculated magnetic moment and the Second postulate of Bohr's model simultaneously to obtain the expression for the magnetic moment of the hydrogen atom.

Formula used:
The formula used in this problem is given as: -
Magnetic moment is,
$M = IA$
where $A$ is the area of the orbit and $I$ is the current.
Bohr’s second postulate,
${m_e}{v_n}{r_n} = \dfrac{{nh}}{{2\pi }}$
where, ${m_e}$ is the mass of an electron, ${r_n}$ is the radius of ${n^{th}}$ Bohr’s orbit, ${v_n}$ is the linear velocity of the electron in ${n^{th}}$ orbit, $h$ is Planck’s constant and $n$ is the principal quantum number.

Complete step by step solution:
It is given that the electron revolves around the nucleus in a radius $r$ with orbital speed $v$. The orbit is equivalent to the magnetic shell of the magnetic moment
$M = IA\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\,(1)$

Now, current $I$ is given by: -
$I = \dfrac{e}{T} = \dfrac{{ev}}{{2\pi r}}$........$\left( {\because T = \dfrac{{2\pi r}}{v}} \right)$
and area of the orbit, $A = \pi {r^2}$
On putting the values of $I$ and $A$ in equation $(1)$, we get
$M = \left( {\dfrac{{ev}}{{2\pi r}}} \right)\pi {r^2}$
$ \Rightarrow M = \dfrac{{evr}}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\,(2)$ proved!

Now, we will determine the expression of the hydrogen atom's magnetic moment using Bohr's theory of quantization. From Bohr’s second postulate of quantization of angular momentum, we get mathematically,
${m_e}{v_n}{r_n} = \dfrac{{nh}}{{2\pi }}$
Substituting ${v_n} = v\,\,and\,\,{r_n} = r$ from the question, we get
$ \Rightarrow {m_e}vr = \dfrac{{nh}}{{2\pi }} \\ $
$ \Rightarrow vr = \dfrac{{nh}}{{2\pi {m_e}}} \\ $
Thus, using the equation $(2)$ , we get the magnetic moment of the hydrogen atom as: -
$ \Rightarrow M = \dfrac{e}{2} \cdot \left( {\dfrac{{nh}}{{2\pi {m_e}}}} \right) = \dfrac{{neh}}{{4\pi {m_e}}} \\ $
For ground state, $n = 1$
$\therefore M = \dfrac{{\left( 1 \right)eh}}{{4\pi {m_e}}} = \dfrac{{eh}}{{4\pi {m_e}}}$

Hence, the expression for the magnetic moment of the hydrogen atom in its ground state, according to Bohr's postulate of the quantization of angular momentum is $M = \dfrac{{eh}}{{4\pi {m_e}}}$.

Note: In this problem, to prove that the electron revolving around the nucleus has a magnetic moment of $evr/2$, we have to analyze all the given conditions and carefully substitute them in $M = IA$ and then the expression for the magnetic moment of the hydrogen atom is obtained by using Bohr’s second postulate. Also, the key point $n = 1$ (for ground state) must be kept in mind while doing the calculation part.