Chapter 28, Problem 21P

To determine

The possible values for $n,l,m_s$

Answer to Problem 21P

The principal quantum number: $n\ge l+1$

The orbital quantum number: $l\ge 3$

The spin quantum number: $m_s=-\frac{1}{2},+\frac{1}{2}$

Explanation of Solution

Given info: The magnetic quantum number: $m_l=-3$

Formula used:

The principal quantum number $n$ can take on any integer value $\left(1,2,3,\mathrm{...}\right)$ and also, corresponds to the quantum number of the old Bohr theory.

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

The magnetic quantum number, $m_l$ , can take on the integer values from $-l$ to $+l$

The spin quantum number, $m_s$ , can be $+\frac{1}{2}$ to $-\frac{1}{2}$

Calculation:

For each $l$ the value of $m_l$ can range from $-l$ to $+l$ and it having that $l\ge 3$

The value of $l$ can range from $\text{0}$ to $\text{n-1}$ .So, it is having that $n\ge l+1$ (minimum four)

For each of these there are two values of $m_s$ : $m_s=-\frac{1}{2},+\frac{1}{2}$

Conclusion:

All the required values are calculated.