Chapter 28, Problem 51GP

To determine

The number of different electron states possible for a given value of $n$ and $\text{2}{n}^2$ .

Explanation of Solution

Given info:

The given value of $l$ is equal 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 $n-1$ .

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

Calculation:

For a given $n,0\le l\le n-1$ .

Since for each $l$ the number of possible states is $2\left(2l+1\right)$ , the number of possible states for a given $n$ is

  ${\displaystyle \sum _{l=0}^{n-1}2\left(2l+1\right)=4{\displaystyle \sum _{l-0}^{n-1}l}}+{\displaystyle \sum _{l-0}^{n-1}2=4\left(\frac{n\left(n-1\right)}{2}\right)}+2n=2n^2$