Question

The orbital angular momentum of an electron corresponding to $n = 4$ and $m = - 3$ is:
A. $0$
B. $\dfrac{h}{{\sqrt 2 \pi }}$
C. $\dfrac{{\sqrt 6 }}{{2\pi }}h$
D. $\dfrac{{\sqrt 3 }}{\pi }h$

Answer 1

Hint: The azimuthal quantum number can be found from the principal quantum number as the value of azimuthal quantum number ranges from $0 \to n - 1$ and the magnetic quantum number is related as $m = + l....0.... - l$ .

Formula used: $L = \sqrt {l\left( {l + 1} \right)} \dfrac{h}{{2\pi }}$
Here, $L$ is the orbital angular momentum, $l$ is the azimuthal quantum number and $h$ is the Planck's constant.

Complete step by step answer:
The orbital angular momentum can be found from the azimuthal quantum number. The azimuthal quantum number can be defined as the quantum number that can be used to determine the shape of the orbital. It is denoted by the alphabet is $l$ . this has a relation with the principal quantum number that is denoted by $n$ and the magnetic quantum number also denoted by $m$ . The principal quantum number can be defined as the quantum number that is used to define the shell to which the electron is assigned, the energy of the orbital. The magnetic quantum number can be defined as the number that is used to determine the orientation of the orbitals. The relation between principal quantum number and azimuthal quantum number is $l = n - 1$ . This implies that the azimuthal quantum number can also be equal to $0$ .
The magnetic quantum number is related to azimuthal quantum number as,
$m = - l \to + l$ this means that the magnetic quantum number can take a value equal to positive $l$ or negative $l$ including $0$.
Using this information, we can find the orbital angular momentum in the steps given below:
$L = \sqrt {l\left( {l + 1} \right)} \dfrac{h}{{2\pi }}$
Here, $L$ is the orbital angular momentum, $l$ is the azimuthal quantum number and $h$ is the plancks constant.
We can find the azimuthal quantum numbers using the information mentioned above. The orbital that contains the $n = 4$ and $m = - 3$ will be found using the following steps:
$l = n - 1$
$ \Rightarrow 4 - 1$
$ \Rightarrow 3$
Therefore, the azimuthal quantum number will be $3$ . therefore, plugging in the value as given, we get,
$L = \sqrt {l\left( {l + 1} \right)} \dfrac{h}{{2\pi }}$
$L = \sqrt {3\left( 4 \right)} \dfrac{h}{{2\pi }}$
$ \Rightarrow \sqrt {12} \dfrac{h}{{2\pi }}$
Taking, the $2$ from the denominator into the root, we get,
$\sqrt {\dfrac{{12}}{4}} \dfrac{h}{\pi }$
$ \Rightarrow \sqrt 3 \dfrac{h}{\pi }$

Therefore, the solution to the question will be, option d that is, angular orbital momentum is $\sqrt 3 \dfrac{h}{\pi }$ .

Note: The principal quantum number is used to determine energy of orbitals and the shell to which the electron is present. The azimuthal quantum number determines the orbital to which it belongs. Magnetic quantum numbers are only for orientation of the orbital.
The magnetic quantum number as well as azimuthal number can take the value of zero. Principal quantum number cannot.