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Hydrogen $(H)$, deuterium $(D)$, singly ionized helium $(He^+)$ and doubly ionized lithium $(Li^{++})$ all have one electron around the nucleus. Consider $n = 2$ to $n = 1$ transition. The wavelengths of emitted radiations are $\lambda_1, \lambda_2 \lambda_3$ and $\lambda_4$ respectively.
$\lambda_1 = \lambda_2 = 4 \lambda_3 = 9 \lambda_4$
$4 \lambda_1 = 2 \lambda_2 = 2 \lambda_3 = \lambda_4$
$\lambda_1 = 2 \lambda_2 = 2 \sqrt 2 \lambda_3 = 3 \sqrt 2 \lambda_4$
$\lambda_1 = \lambda_2 = 2 \lambda_3 = 3 \sqrt 2 \lambda_4$
Solution
$Z_{1}=1, Z_{2}=1, Z_{3}=2$ and $Z_{4}=3$
$\frac{1}{\lambda}=R Z^{2}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)$ or $\quad \lambda=\frac{4}{3 R Z^{2}}$
or $\quad \lambda Z^{2}=$ constant
So $\lambda_{1}(1)^{2}=\lambda_{2}(1)^{2}=\lambda_{3}(2)^{2}=\lambda a\left(3^{2}\right)$
or $\lambda_{1}=\lambda_{1} \lambda_{2}=4 \lambda_{3}=9 \lambda_{4}$
