Q73P - Fundamentals Of Physics

Question

If an electron in an atom has orbital angular momentum with values limited by 3, how many values of (a) ${\mathit{L}}_{\mathbf{o}\mathbf{r}\mathbf{b}\mathbf{,}\mathbf{z}}$ and (b) ${\mathit{\mu }}_{\mathbf{o}\mathbf{r}\mathbf{b}\mathbf{,}\mathbf{z}}$ can the electron have? In terms of h, m, and e, what is the greatest allowed magnitude for (c) ${\mathit{L}}_{\mathbf{o}\mathbf{r}\mathbf{b}\mathbf{,}\mathbf{z}}$ and (d) ${\mathit{\mu }}_{\mathbf{o}\mathbf{r}\mathbf{b}\mathbf{,}\mathbf{z}}$? (e) What is the greatest allowed magnitude for the z component of the electron’s net angular momentum (orbital plus spin)? (f) How many values (signs included) are allowed for the z component of its net angular momentum?

Step-by-Step Solution

Answer

(a) Number of values of $L_{orb,z}$ that an electron can have is seven.

(b) Number of values of ${\mu }_{orb,z}$ that an electron can have is seven.

(c) Greatest allowed magnitude of $L_{orb,z}$ is $\frac{3h}{2\pi }$.

(d) Greatest allowed magnitude of ${\mu }_{orb,z}$ is $\frac{3eh}{4\pi m_e}$.

(e) Greatest allowed magnitude for z component of the electron’s net angular momentum is $\frac{3.5h}{2\pi }$.

(f) Number of values allowed to magnitude for z component of the electron’s net angular momentum is 8.

Understanding the concepts of magnetic field:

By using the concept of quantum numbers, we can find the number of values and greatest allowed values.

Formulae:

Number of different values of  $m_l$ is given by $(2l+1)$.

The angular momentum is,

$L_{orb,z}=\frac{m_{l}h}{2\pi }$

(a) Calculations of Number of values of L o r b , z :

For the given value of $l, m_l$ varies from $–l$ to $l$. Thus, in this case, $l=3$, and the number of different values of $m_l$ is

$2l+1=2\left(3\right)+1=7$

So, there are seven different values of  $L_{orb,z}$.

(b) Calculations of number of values of μ o r b , z :

Similarly, since ${\mu }_{orb,z}$ is directly proportional to $m_l$, there are total seven different values of ${\mu }_{orb,z}$.

(c) Calculations of greatest allowed magnitude of L o r b , z :

As you know,  

$L_{orb,z}=\frac{m_{l}h}{2\pi }$

So, the greatest allowed value of  $L_{orb,z}$ is given by

$\frac{{\left|m_l\right|}_{maximum}h}{2\pi }=\frac{3h}{2\pi }$

Greatest allowed magnitude of $L_{orb,z}$ is $\frac{3h}{2\pi }$.

(d) Calculations of greatest allowed magnitude of μ o r b , z :

Since  ${\mu }_{orb,z}=-m_l\mu B$, the greatest allowed value of ${\mu }_{orb,z}$ is given by

${\left|m_l\right|}_{maximum}\mu B=\frac{3eh}{4\pi m_e}$

Greatest allowed magnitude of ${\mu }_{orb,z}$ is $\frac{3eh}{4\pi m_e}$

(e) Calculations of greatest allowed magnitude for z component of the electron’s net angular momentum:

The z component of the net angular momentum of the electron is given by 

$\begin{array}{rcl}{L}_{net,z}& =& L_{orb,Z}+L_{s,Z}\\ & =& \frac{m_{l}h}{2\pi }+\frac{m_{s}h}{2\pi }\end{array}$

From the given value,  Thus, $m_l=3$ and $m_s=\frac{1}{2}$.

$\begin{array}{rcl}{\left[L_{net,z}\right]}_{maximum}& =& \frac{\left(3+\frac{1}{2}\right)h}{2\pi }\\ & =& \frac{3.5h}{2\pi }\end{array}$

Hence. the greatest allowed magnitude for z component of the electron’s net angular momentum is $\frac{3.5h}{2\pi }$.