Chapter 28, Problem 43GP

To determine

The largest and smallest possible values for the angular momentum of an electron.

Answer to Problem 43GP

The largest and smallest possible values are $\text{4}\text{.72}{\text{×10}}^{\text{-34}}\text{kg}\cdot {\text{m}}^{\text{2}}\text{/s}$ and zero respectively.

Explanation of Solution

Given info:

Number, $\text{n=5}$

Formula used:

The orbital quantum number $\mathrm{l}$ can take on values from $0$ to $\text{n-1}$ .

The actual magnitude of the angular momentum $\text{L}$ is related to the quantum $\mathrm{l}$ by

  $\text{L = }\mathrm{\hslash }{\left[\mathrm{l}\left(\mathrm{l}+1\right)\right]}^{\frac{1}{2}}$

Where, $\mathrm{l}$ is the orbital quantum number and $\mathrm{\hslash }$ is reduced Planck’s constant, $1.055\times {10}^{-34}\mathrm{J}\cdot \mathrm{s}$

Calculation:

The value of $\mathrm{l}$ can range from $0$ to $\text{n-1}$ . Therefore, for $\text{n}\text{=}\text{5}$ the $\mathrm{l}$ will be: $\mathrm{l}\le 4$

So, the smallest value of $\text{L}$ is zero.

The largest magnitude of $\text{L}$ :

  $\begin{array}{l}\text{L = }\mathrm{\hslash }{\left[\mathrm{l}\left(\mathrm{l}+1\right)\right]}^{\frac{1}{2}}=\left(\text{1}\text{.055}\text{×}{\text{10}}^{\text{-34}}\text{J}\cdot \text{s}\right){\left[\left(\text{4}\right)\left(\text{4+1}\right)\right]}^{\frac{\text{1}}{\text{2}}}\\ \text{= 4}\text{.72}{\text{×10}}^{\text{-34}}\text{kg }\cdot {\text{ m}}^{\text{2}}\text{/s}\end{array}$

Conclusion:

The largest and smallest possible values are $\text{4}\text{.72}{\text{×10}}^{\text{-34}}\text{kg}\cdot {\text{m}}^{\text{2}}\text{/s}$ and zero respectively.