Question

A hydrogen atom is in a state that has ${\mathit{L}}_{\mathbf{z}}\mathbf{=}\mathbf{2}\sout{\mathbf{h}}$In the semi classical vector model, the angular momentum vector $\overrightarrow{\mathbf{L}}$for this state makes an angle${\mathbf{\theta }}_{\mathbf{L}}\mathbf{=}\mathbf{63}\mathbf{.}\mathbf{4}\mathbf{°}$with the +z-axis. (a) What is the l quantum number for this state? (b) What is the smallest possible n quantum number for this state?

Short answer

1. The value of the orbital quantum number is l=4.
2. The smallest possible value of n is 5

Step-by-step solution

Important Concepts

Orbital angular momentum in a hydrogen atom is given by

$\mathbf{L}\mathbf{=}\sqrt{\mathbf{l}\mathbf{+}\mathbf{1}}\sout{\mathbf{h}}$………………(1)

Where l is the azimuthal quantum number.

The angle the vector L makes with the z-axis is given by

$\mathbf{\theta }\mathbf{=}{\mathbf{cos}}^{\mathbf{-}\mathbf{1}}\frac{{\mathbf{L}}_{\mathbf{z}}}{\mathbf{L}}$……………….(2)

Application

Solving the equation (2) for we get,

$\mathrm{L}=\frac{{\mathrm{L}}_{\mathrm{z}}}{\mathrm{cos\theta }}$

Plug in the given values

$\begin{array}{l}\mathrm{L}=\frac{2\sout{\mathrm{h}}}{\cos \left(63.4^0\right)}\\ \mathrm{L}=4.47\sout{\mathrm{h}}\end{array}$

Square both sides of equation (1) and rearranging it we get,

$$\begin{gathered} {\mathrm{L}}^2=\mathrm{l}\left(\mathrm{l}+1\right){\sout{\mathrm{h}}}^2 \\ {\mathrm{l}}^2+\mathrm{l}-\frac{{\mathrm{L}}^2}{{\sout{\mathrm{h}}}^2}=0 \\ {\mathrm{l}}^2+\mathrm{l}-20=0 \end{gathered}$$

Solve for l and we get

l = 4 and l = - 5

Reject the negative value, therefore the orbital quantum number of this state is l=4

Since orbital quantum numbers can have values from 0 to n - 1

So the possible values of n for this state is $\mathrm{n}\ge \mathrm{l}+1$

Hence$\mathrm{n}\ge 5$

Hence the minimum value of n is 5