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Question: An electron is in a state with \(l = 3\) (a) what is multiple of \(\hbar \) gives the magnitude of \...

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

Explanation

Solution

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 2π\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,ml,sn,l,{m_l},s. Here, nn is known as principal quantum number, ll is azimuthal quantum number, ml{m_l} is known as magnetic quantum number and, ss is known as spin quantum number.

For (a) since given l=3l = 3 so according to formula:
L=l(l+1)\left| {\vec L} \right| = \sqrt {l(l + 1)} \hbar
Here, L\vec L is known as orbital angular momentum, so putting values we get, L=3(3+1)\left| {\vec L} \right| = \sqrt {3(3 + 1)} \hbar
L=12\left| {\vec L} \right| = \sqrt {12} \hbar
L=3.46\Rightarrow \left| {\vec L} \right| = 3.46\hbar
So a multiple of \hbar is 3.463.46 gives the magnitude of L\vec L

For (b) part, Let, μb{\mu _b} be a radial magnetic dipole moment and μ\mu be a magnetic dipole moment. Since we know, μ=l(l+1)μb\mu = \sqrt {l(l + 1)} {\mu _b}
Since, l=3l = 3 so putting values we get,
μ=3(3+1)μb\mu = \sqrt {3(3 + 1)} {\mu _b}
μ=12μb\Rightarrow \mu = \sqrt {12} {\mu _b}
μ=3.46μb\therefore \mu = 3.46{\mu _b}
So multiple of μB{\mu _B} gives the magnitude of μ\vec \mu is 3.463.46

For (C) part, since we know ml{m_l} varies from l - l to +l + l
Since l=3l = 3 so ml{m_l} have values, 3,2,1,0,1,2,3 - 3, - 2, - 1,0,1,2,3
So the largest value of ml{m_l} is 33.

For (d) part, let Lz{L_z} be orbital angular momentum in z axis, so from formula;
Lz=ml{L_z} = {m_l}\hbar
Since ml{m_l} has the largest value 33.
So, Lz=3{L_z} = 3\hbar
Multiple of \hbar gives the corresponding value of Lz{L_z} is 33.

For (e) part, let μorb,z{\mu _{orb,z}} be orbital magnetic dipole moment, so according to formula; μorb,z=mlμb{\mu _{orb,z}} = - {m_l}{\mu _b}
Since largest value of ml{m_l} is 33 so, μorb,z=3μb{\mu _{orb,z}} = - 3{\mu _b}
Multiple of μB{\mu _B} gives the corresponding value of μorb,z{\mu _{orb,z}} is 3 - 3

For (f) part, let θ\theta be semi classical angle between L\vec L and Lz{L_z} so from formula it is defined as, cosθ=mll(l+1)\cos \theta = \dfrac{{{m_l}}}{{\sqrt {l(l + 1)} }}
So for l=3&ml=3l = 3\,\& \,{m_l} = 3 (Since 33 is largest value of ml{m_l} ) we have,
cosθ=33(3+1)\cos \theta = \dfrac{3}{{\sqrt {3(3 + 1)} }}
cosθ=312\Rightarrow \cos \theta = \dfrac{3}{{\sqrt {12} }}
cosθ=32\Rightarrow \cos \theta = \dfrac{{\sqrt 3 }}{2}
θ=30\therefore \theta = 30^\circ

For (g) part, second largest value so for, l=3&ml=2l = 3\,\&\, {m_l} = 2 we have,
cosθ=23(3+1)\cos \theta = \dfrac{2}{{\sqrt {3(3 + 1)} }}
cosθ=212\Rightarrow \cos \theta = \dfrac{2}{{\sqrt {12} }}
cosθ=13\Rightarrow \cos \theta = \dfrac{1}{{\sqrt 3 }}
θ=54.7\therefore \theta = 54.7^\circ

For (h) part, cosθ=33(3+1)\cos \theta = \dfrac{{ - 3}}{{\sqrt {3(3 + 1)} }} since ( l=3&ml=3l = 3\,\&\, {m_l} = - 3 )
cosθ=312\Rightarrow\cos \theta = \dfrac{{ - 3}}{{\sqrt {12} }}
cosθ=32\Rightarrow \cos \theta = \dfrac{{ - \sqrt 3 }}{2}
θ=150\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.