Answer:
The answer is "[tex]32819.9 \ \frac{J}{mol}\\\\[/tex]"
Explanation:
[tex]Boron: 5^{B}\to 1s^2 2s^2 2p^1[/tex]
[tex]\Delta E=-2.18\times 10^{-18}\ \frac{J}{atom}\ (\frac{1}{\infity^2}-\frac{1}{n^2_{initial}})(z^2) (6.022\times 10^{23}\ \frac{atom}{mol})\\\\[/tex]
[tex]=-2.18\times 10^{-18}\ \frac{J}{atom}\ (0-\frac{1}{1})(5^2) (6.022\times 10^{23}\ \frac{atom}{mol})\\\\ =2.18\times 10^{-18}\times 25 \times 6.022\times 10^{23}\ (\frac{J}{mol})\\\\ =328.199 \times 10^{5}\ (\frac{J}{mol})\\\\ =32819 \times 10^{3}\ (\frac{J}{mol})\\\\ =32819.9 \ (\frac{J}{mol})\\\\[/tex]