Explanation

(b)
$\mathrm{\Delta}E = hv = \frac{2\pi^{2}mZ^{2}e^{4}k^{2}}{h^{2}}\left\lbrack \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right\rbrack$
If electron falls from $n_{2} -$level to $n_{1} -$level.
$\therefore$ In $He^{+}$ for the $n_{2} = 4\ to\ n_{1} = 2$ transition
$\upsilon\left( He^{+} \right) = constant\ (4)\left\lbrack \frac{1}{2^{2}} - \frac{1}{4^{2}} \right\rbrack\ \ \ \ \ \left\lbrack \because Z_{He^{+}} = 2 \right\rbrack$
$= constant\ \times 4\ \left\lbrack \frac{3}{16} \right\rbrack = \frac{3}{4}constant$
$\upsilon(H) = constant\ (1)^{2}\left\lbrack \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right\rbrack$
$= constant\ \times \left\lbrack \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right\rbrack\$
$(a)For\ n_{2} = 3\ and\ n_{1} = 1,\$
$\upsilon(H) = constant\left\lbrack \frac{1}{1} - \frac{1}{9} \right\rbrack$
$= \frac{8}{9}constant\$
$\neq \frac{3}{4} \times constant\$
$(b)For\ n_{2} = 2\ and\ n_{1} = 1,\$
$\upsilon(H) = \ constant\ \times \left\lbrack \frac{1}{1} - \frac{1}{4} \right\rbrack$
$= \frac{3}{4} \times constant$
$= \upsilon\ \left( He^{+} \right)$