Explanation

(b) From Bohr’s formula , the wave number $\left( \frac{1}{\lambda} \right)$ is given by
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{\lambda}$ $= \ Z^{2}R\ \left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{\ \ \ 2}} \right)$
where $Z\ $is atomic number, $R\ $the Rydberg’s constant and $n$ the quantum number.
$\Rightarrow \ \ \ \ \ \ \ \ \ \lambda \propto \frac{1}{Z^{2}}$
Atomic number of lithium is 3, of helium is 2 and of hydrogen is 1.
$\therefore\ \ \ \ \ \ \ \ \ \ \lambda_{{Li}^{2 +}}:\lambda_{{He}^{+}}:\lambda_{H}$=$\frac{1}{(3)^{2}}:\frac{1}{(2)^{2}}:1$
=$\frac{1}{9}$ :$\frac{1}{4}$ :1