Quantum Chemistry/Example 26

The square of the angular momentum of a hydrogen atom is measured to be $$L^2 $$$$= 20\hbar^2 $$. What are the possible values of the z-component of the orbital angular momentum, $$L_z $$, that could be measured for this atom?

 Solution: 

The eigenvalues of the square of the angular momentum operator ($$\hat{L}$$2) for a quantum mechanical system, such as an electron in a hydrogen atom, are given by:

$$\hat{L}^2 Y_l^m (\theta,\phi)= h^2 l (l + 1) Y_l^m (\theta,\phi) $$

where $$Y_l^m (\theta,\phi) $$ are the spherical harmonics, which are eigenfunctions of $$\hat{L}$$2, $$l $$ is the orbital quantum number, and $$m $$ is the magnetic quantum number. The given value for $$L^2 $$ = 20 ħ2, so we set up the equation:

$$\hbar^2l(l+1)=20\hbar^2 $$

Dividing by $$\hbar^2 $$ and simplifying, we get:

$$l(l+1)=20 $$

$$l^2+l-20=0 $$

In this quadratic equation, $$l $$ can be factored to get:

$$(l-4)(l+5)=0 $$

$$l=4 $$

Since $$l $$ must be a non-negative integer. The magnetic quantum number $$m $$ can take on any integer value from $$-l $$ to $$l $$, thus for $$l=4 $$, $$m $$ can be:

$$m=-4, -3, -2, -1, 0, 1, 2, 3, 4 $$

The z-component of the angular momentum, $$L_z $$, is quantized in units of $$\hbar^2 $$and given by:

$$L_z=m\hbar $$

$$L_z=-4\hbar, -3\hbar, -2\hbar, -1\hbar, 0\hbar, 1\hbar, 2\hbar, 3\hbar, 4\hbar $$.

Therefore, the possible values of the z-component of the orbital angular momentum, $$L_z $$, that could be measured for the atom with a given $$L^2 $$$$= 20\hbar^2 $$ are $$L_z=-4\hbar, -3\hbar, -2\hbar, -1\hbar, 0\hbar, 1\hbar, 2\hbar, 3\hbar, 4\hbar  $$.