Chapter 28, Problem 16P

To determine

The quantum numbers for each given state.

Answer to Problem 16P

The answer is $32states$

Explanation of Solution

Given info: The principal quantum number $\text{n = 4}$

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:

The value of $l$ can range from $\text{0}$ to $\text{n-1}$ . So, for $\text{n = 4}$ , we have $l=0,1,2,3$ .

For each $l$ the value of $m_l$ can range from $-l$ to $+l$ , or $2l+1$ values.

Also, for each of these there are two values of $m_s$ .

So, the sum number of each $l$ is $2\left(2l+1\right)$ .

The number of states is $N=2\left(0+1\right)+2\left(2+1\right)+2\left(4+1\right)+2\left(6+1\right)=32states$

Now, start with $l=0$ , and also, list quantum numbers in order $\left(n,l,m_l,m_s\right)$ ;

  $\begin{array}{l}\left(4,0,0,-\frac{1}{2}\right),\left(4,0,0,+\frac{1}{2}\right),\left(4,1,-1,-\frac{1}{2}\right),\left(4,1,-1,+\frac{1}{2}\right),\left(4,1,0,-\frac{1}{2}\right),\\ \left(4,1,0,+\frac{1}{2}\right),\left(4,1,1,-\frac{1}{2}\right),\left(4,1,1,+\frac{1}{2}\right),\left(4,2,-2,-\frac{1}{2}\right),\left(4,2,-2,+\frac{1}{2}\right),\\ \left(4,2,-1,-\frac{1}{2}\right),\left(4,2,-1,+\frac{1}{2}\right),\left(4,2,0,-\frac{1}{2}\right),\left(4,2,0,+\frac{1}{2}\right),\left(4,2,1,-\frac{1}{2}\right),\\ \left(4,2,1,+\frac{1}{2}\right),\left(4,2,2,-\frac{1}{2}\right),\left(4,2,2,+\frac{1}{2}\right),\left(4,3,-3,-\frac{1}{2}\right),\left(4,3,-3,+\frac{1}{2}\right),\\ \left(4,3,-2,-\frac{1}{2}\right),\left(4,3,-2,+\frac{1}{2}\right),\left(4,3,-1,-\frac{1}{2}\right),\left(4,3,-1,+\frac{1}{2}\right),\left(4,3,0,-\frac{1}{2}\right),\\ \left(4,3,0,+\frac{1}{2}\right),\left(4,3,1,-\frac{1}{2}\right),\left(4,3,1,+\frac{1}{2}\right),\left(4,3,2,-\frac{1}{2}\right),\left(4,3,2,+\frac{1}{2}\right),\\ \left(4,3,3,-\frac{1}{2}\right),\left(4,3,3,+\frac{1}{2}\right)\end{array}$

Conclusion:

The required value is $32states$