Question

A particle orbiting due to an attractive central force has angular momentum $\mathrm{L}=1.00\times {10}^{-33}\mathrm{kg}.\mathrm{m}/\mathrm{s}$ What z-components of angular momentum is it possible to detect?

Short answer

The possible z-components of angular momentum are,

${\mathrm{L}}_{\mathrm{z}}+0,\pm \mathrm{h},\pm 2\mathrm{h},\pm 3\mathrm{h},\pm 4\mathrm{h},\pm 5\mathrm{h},\pm 6\mathrm{h}.\pm 7\mathrm{h},\pm 8\mathrm{h},\pm 9\mathrm{h}$

Step-by-step solution

A concept:

The azimuthal quantum number specifies the shape and angular momentum of the orbital in space.

Getting the value of I :

Consider the given data as below.

Angular momentum, $\mathrm{L}=1.00\times {10}^{-33}\mathrm{kg}.\mathrm{m}/\mathrm{s}$

The angular momentum is given by,

$\mathrm{L}=\sqrt{\mathrm{I}\left(\mathrm{I}+1\right)}\times \mathrm{h}$ ….. (1)

Where, I is the Azimuthal Quantum number, h is Plank’s constant that is $1.055\times {10}^{-34}\mathrm{J}.\mathrm{s}.$

$$\begin{gathered} 9.478=\sqrt{\mathrm{I}\left(\mathrm{I}+1\right)} \\ {\mathrm{I}}^2+\mathrm{I}-89.832=0 \end{gathered}$$

Solving above equation gives you get,

I = 9

z-components of angular momentum:

As you know that, z-component to angular momentum,

${\mathrm{L}}_{\mathrm{z}}={\mathrm{m}}_{\mathrm{l}}\mathrm{h}$

Where, ${\mathrm{m}}_{\mathrm{l}}$is the magnetic quantum number that is$0,\pm 1,\pm 2,...\pm \mathrm{I}$

Hence, possible z-components are,

${\mathrm{L}}_{\mathrm{z}}+0,\pm \mathrm{h},\pm 2\mathrm{h},\pm 3\mathrm{h},\pm 4\mathrm{h},\pm 5\mathrm{h},\pm 6\mathrm{h}.\pm 7\mathrm{h},\pm 8\mathrm{h},\pm 9\mathrm{h}$