Chapter 28, Problem 50GP

(a)

To determine

The number of different states for a given value of $l$ is equals to $2\left(2l+1\right)$

Answer to Problem 50GP

The given condition is shown.

Explanation of Solution

Given info: The given value of $l$ is equals to $2\left(2l+1\right)$

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 $+l$

Calculation:

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

For each of these there are two values of $m_s$ .

So, the total number of states in a sub shell is $\text{N = 2}\left(\text{2}l\text{+1}\right)$ .

(b)

To determine

The number for $l=0,1,2,3,4,5,6$

Answer to Problem 50GP

The number for $l=0,1,2,3,4,5,6$ is $\text{N =}\text{2,6,10,14,18,22,26}$ respectively.

Explanation of Solution

Given info: The given value of $l$ is equals to $2\left(2l+1\right)$

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 $+l$

Calculation:

According to the rule, for orbital quantum number $l=0,1,2,3,4,5,6$ is $\text{N =}\text{2,6,10,14,18,22,26}$ respectively.