Chapter 36, Problem 32P

(a)

To determine

The total number of electron state for the given principal quantum number.

Answer to Problem 32P

The total number of electron state is $32$ .

Explanation of Solution

Given:

The principal quantum number is $n=4$ .

Formula used:

The expression to calculate the possible values of orbital number is given by,

  $l=x,\left(0\le x<n\right)$

Here $x$ is the integer.

The expression to calculate the possible values of magnetic orbital quantum number is given by,

  $m=x,\left(-l_{\text{n}}\le x\le l_{\text{n}}\right)$

Here $x$ is the integer.

Calculation:

The possible values of the orbital number is calculated as,

  $l=x,\left(0\le x<4\right)$

So, the possible value of the orbital numbers are $0,1,2,3$ .

The possible values of the magnetic orbital quantum number for $l=0$ is calculated as,

  $m=x,\left(0\le x\le 0\right)$

So, the possible value of the magnetic orbital numbers is $0$ .

The possible values of the magnetic orbital quantum number for $l=1$ is calculated as,

  $m_{\text{l}}=x,\left(-1\le x\le 1\right)$

So, the possible value of the magnetic orbital numbers is $-1,0,1$ .

The possible values of the magnetic orbital quantum number for $l=2$ is calculated as,

  $m_{\text{l}}=x,\left(-2\le x\le 2\right)$

So, the possible value of the magnetic orbital numbers is $-2,-1,0,1,2$ .

The possible values of the magnetic orbital quantum number for $l=3$ is calculated as,

  $m_{\text{l}}=x,\left(-3\le x\le 3\right)$

So, the possible value of the magnetic orbital numbers is $-3,-2,-1,0,1,2,3$ .

The different possible combinations of $\left(n,l,m_{\text{l}}\right)$ for the given state are tabulated below.

$n$	$l$	$m_{\text{l}}$
$4$	$0$	$0$
$4$	$1$	$-1$
$4$	$1$	$0$
$4$	$1$	$1$
$4$	$2$	$-2$
$4$	$2$	$-1$
$4$	$2$	$0$
$4$	$2$	$1$
$4$	$2$	$2$
$4$	$3$	$-3$
$4$	$3$	$-2$
$4$	$3$	$-1$
$4$	$3$	$0$
$4$	$3$	$1$
$4$	$3$	$2$
$4$	$3$	$3$

Table 1

The possible number of combination is $16$ as the electron has two possible spins and the number of electronic states is $32$ .

Conclusion:

Therefore, the total number of electron states is $32$ .

(b)

To determine

The total number of electron state for the given principal quantum number.

Answer to Problem 32P

The total number of electron state is $8$ .

Explanation of Solution

Given:

The principal quantum number is $n=2$ .

Formula used:

The expression to calculate the possible values of orbital number is given by,

  $l=x,\left(0\le x<n\right)$

Here $x$ is the integer.

The expression to calculate the possible values of magnetic orbital quantum number s given by,

  $m=x,\left(-l_{\text{n}}\le x\le l_{\text{n}}\right)$

Here $x$ is the integer.

Calculation:

The possible values of the orbital number is calculated as,

  $l=x,\left(0\le x<2\right)$

So, the possible value of the orbital numbers are $0,1$ .

The possible values of the magnetic orbital quantum number for $l=0$ is calculated as,

  $m=x,\left(0\le x\le 0\right)$

So, the possible value of the magnetic orbital numbers is $0$ .

The possible values of the magnetic orbital quantum number for $l=1$ is calculated as,

  $m_{\text{l}}=x,\left(-1\le x\le 1\right)$

So, the possible value of the magnetic orbital numbers is $-1,0,1$ .

The different possible combinations of $\left(n,l,m_{\text{l}}\right)$ for the given state are tabulated below.

$n$	$l$	$m_{\text{l}}$
$2$	$0$	$0$
$2$	$1$	$-1$
$2$	$1$	$0$
$2$	$1$	$1$

Table 2

The possible number of combination is $4$ as the electron has two possible spins and the number of electronic states is $8$ .

Conclusion:

Therefore, the total number of electron states is $8$ .