Chapter 28, Problem 53GP

To determine

To Compare: The results for $\text{n = 2}$

Explanation of Solution

Given info:

Number, $\text{n}\text{=}\text{2}$

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:

According to the Bohr model,

  ${\text{L}}_{\text{Bohr}}\text{= n}\frac{\text{h}}{\text{2π}}\text{=}\text{2}\mathrm{\hslash }$

According to the quantum mechanics,

  ${\text{L}}_{\text{QM}}\text{=}\sqrt{\text{l}\left(\text{l + 1}\right)}{\mathrm{\hslash }}_{\text{p}}$ and with $\text{n - 2}$ , it follows that $\mathrm{l}=0$ or $\mathrm{l}=1$

So, that ${\text{L}}_{\text{QM}}\text{=}\sqrt{\text{0}\left(\text{0 + 1}\right)}\mathrm{\hslash }=0$ or ${\text{L}}_{\text{QM}}\text{=}\sqrt{\text{1}\left(\text{1 + 1}\right)}\mathrm{\hslash }=\sqrt{\text{2}}\mathrm{\hslash }$