Question

An electron is in a state with $l = 3$ (a) what is multiple of $\hbar $ gives the magnitude of $\vec L$ ? (b) What multiple of ${\mu _B}$ gives the magnitude of $\vec \mu $ ? (c) What is largest possible value of ${m_l}$ ? (d) What multiple of $\hbar $ gives the corresponding value of ${L_z}$ and (e) what multiple of ${\mu _B}$ gives the corresponding value of ${\mu _{orb,z}}$ ? (f) What is value of semi classical angle $\theta $ between the directions of ${L_z}$ and $\vec L$ what is value of $\theta $ for (g) the second largest possible value of ${m_l}$ and (h) the smallest (that is, most negative) possible value of ${m_l}$ ?

Answer 1

Hint: In order to solve this question we need to understand the atomic structure of hydrogen atoms. So structure of hydrogen atom was first suggested by Niels Bohr, a Danish physicist, according to him electrons around hydrogen and hydrogen like atoms are in shells having a particular combination, and their angular momentum is integral multiple of $\dfrac{\hbar }{{2\pi }}$ so with the help of his theory and two forces he assumed that the electrons are moving in shells around nucleus, however after quantum mechanics came into existence, which predicts that electrons have a wave function associated with and after calculation it had been found that electrons have basic quantum numbers, which decide in which electron would reside.

Complete step by step answer:
The four basic quantum numbers are, $n,l,{m_l},s$. Here, $n$ is known as principal quantum number, $l$ is azimuthal quantum number, ${m_l}$ is known as magnetic quantum number and, $s$ is known as spin quantum number.

For (a) since given $l = 3$ so according to formula:
$\left| {\vec L} \right| = \sqrt {l(l + 1)} \hbar $
Here, $\vec L$ is known as orbital angular momentum, so putting values we get, \[\left| {\vec L} \right| = \sqrt {3(3 + 1)} \hbar \]
\[\left| {\vec L} \right| = \sqrt {12} \hbar \]
\[\Rightarrow \left| {\vec L} \right| = 3.46\hbar \]
So a multiple of \[\hbar \] is $3.46$ gives the magnitude of $\vec L$

For (b) part, Let, ${\mu _b}$ be a radial magnetic dipole moment and $\mu $ be a magnetic dipole moment. Since we know, $\mu = \sqrt {l(l + 1)} {\mu _b}$
Since, $l = 3$ so putting values we get,
$\mu = \sqrt {3(3 + 1)} {\mu _b}$
$\Rightarrow \mu = \sqrt {12} {\mu _b}$
$\therefore \mu = 3.46{\mu _b}$
So multiple of ${\mu _B}$ gives the magnitude of $\vec \mu $ is $3.46$

For (C) part, since we know ${m_l}$ varies from $ - l$ to $ + l$
Since $l = 3$ so ${m_l}$ have values, $ - 3, - 2, - 1,0,1,2,3$
So the largest value of ${m_l}$ is $3$.

For (d) part, let ${L_z}$ be orbital angular momentum in z axis, so from formula;
${L_z} = {m_l}\hbar $
Since ${m_l}$ has the largest value $3$.
So, ${L_z} = 3\hbar $
Multiple of $\hbar $ gives the corresponding value of ${L_z}$ is $3$.

For (e) part, let ${\mu _{orb,z}}$ be orbital magnetic dipole moment, so according to formula; ${\mu _{orb,z}} = - {m_l}{\mu _b}$
Since largest value of ${m_l}$ is $3$ so, ${\mu _{orb,z}} = - 3{\mu _b}$
Multiple of ${\mu _B}$ gives the corresponding value of ${\mu _{orb,z}}$ is $ - 3$

For (f) part, let $\theta $ be semi classical angle between $\vec L$ and ${L_z}$ so from formula it is defined as, $\cos \theta = \dfrac{{{m_l}}}{{\sqrt {l(l + 1)} }}$
So for $l = 3\,\& \,{m_l} = 3$ (Since $3$ is largest value of ${m_l}$ ) we have,
$\cos \theta = \dfrac{3}{{\sqrt {3(3 + 1)} }}$
$\Rightarrow \cos \theta = \dfrac{3}{{\sqrt {12} }}$
$\Rightarrow \cos \theta = \dfrac{{\sqrt 3 }}{2}$
$\therefore \theta = 30^\circ $

For (g) part, second largest value so for, $l = 3\,\&\, {m_l} = 2$ we have,
$\cos \theta = \dfrac{2}{{\sqrt {3(3 + 1)} }}$
$\Rightarrow \cos \theta = \dfrac{2}{{\sqrt {12} }}$
$\Rightarrow \cos \theta = \dfrac{1}{{\sqrt 3 }}$
$\therefore \theta = 54.7^\circ $

For (h) part, $\cos \theta = \dfrac{{ - 3}}{{\sqrt {3(3 + 1)} }}$ since ( $l = 3\,\&\, {m_l} = - 3$ )
$\Rightarrow\cos \theta = \dfrac{{ - 3}}{{\sqrt {12} }}$
$\Rightarrow \cos \theta = \dfrac{{ - \sqrt 3 }}{2}$
$\therefore \theta = 150^\circ $

Note: It should be remembered that the value of orbital angular momentum, dipole moment, is not defined arbitrarily but they have been derived by solving Schrodinger equation a proceeding with various constraints, also by using special functions. Also Pauli Exclusion Principle states that in an orbital, no two electrons have the same set of quantum numbers.