30.9 The pauli exclusion principle  (Page 5/11)

(a) 32. (b) $2\text{in}s,\text{6 in}p,\text{10 in}d,$ and 14 in $f$ , for a total of 32.

(a) 2

(b) $3d^9$

(b) $n\ge l$ is violated,

(c) cannot have 3 electrons in $s$ subshell since $3>(2l+1)=2$

(d) cannot have 7 electrons in $p$ subshell since $7>(2l+1)=2(2+1)=6$

(a) The number of different values of $m_l$ is $\pm l,\pm (l-1),\text{...,}0$ for each $l>0$ and one for $l=0\Rightarrow (2l+1)\text{.}$ Also an overall factor of 2 since each $m_l$ can have $m_s$ equal to either $+1/2$ or $-1/2\Rightarrow 2(2l+1)$ .

(b) for each value of $l$ , you get $2(2l+1)$

$=\text{0, 1, 2, ...,}(n\text{–1})\Rightarrow 2\left\{\left[(2)(0)+1\right]+\left[(2)(1)+1\right]+\text{.}\text{.}\text{.}\text{.}+\left[(2)(n-1)+1\right]\right\}=\underset{\underset{n\text{terms}}{︸}}{2\left[1+3+\text{.}\text{.}\text{.}+(2n-3)+(2n-1)\right]}$ to see that the expression in the box is $=n^2,$ imagine taking $(n-1)$ from the last term and adding it to first term $=2\left[1+(n\text{–1})+3+\text{.}\text{.}\text{.}+(2n-3)+(2n-1)–(n-1)\right]=2\left[n+3+\text{.}\text{.}\text{.}\text{.}+(2n-3)+n\right]\text{.}$ Now take $(n-3)$ from penultimate term and add to the second term $2\underset{\underset{n\text{terms}}{︸}}{\left[n+n+\text{.}\text{.}\text{.}+n+n\right]}={2n}^2$ .