Explanation

(a) $E = \ Rhc\left\lbrack \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right\rbrack$
$\ \ \ \ \ \ \ \ \ \ \ \ \ {\ E}_{(4 \rightarrow 3)}$ =$Rhc\left\lbrack \frac{1}{3^{2}} - \frac{1}{4^{2}} \right\rbrack$
=$Rhc\left\lbrack \frac{7}{\ 9\ \times \ 16} \right\rbrack$= 0.05 $Rhc$
$\ \ \ \ \ \ \ \ \ \ \ \ \ E_{(4 \rightarrow 2)}$ =$Rhc\left\lbrack \frac{1}{2^{2}} - \frac{1}{4^{2}} \right\rbrack$
=$Rhc\left\lbrack \frac{3}{\ 16} \right\rbrack$= 0.2 Rhc
$\ \ \ \ \ \ \ \ \ \ \ \ E_{(2 \rightarrow 1)}$ =$Rhc\left\lbrack \frac{1}{{(1)}^{2}} - \frac{1}{{(2)}^{2}} \right\rbrack$
=$Rhc\left\lbrack \frac{3}{\ 4} \right\rbrack$= 0.75 $Rhc$
$\ \ \ \ \ \ \ \ \ \ \ \ E_{(1 \rightarrow 3)}$ =$Rhc\left\lbrack \frac{1}{{(3)}^{2}} - \frac{1}{{(1)}^{2}} \right\rbrack$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = -$ $\frac{8}{\ 9\ }\ Rhc\ $= $- 0.9\ Rhc$
Thus, transition III gives most energy. Transition I represents the absorption of energy.