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Q.
The ratio of the wavelengths for $2 \rightarrow 1$ transition in $Li^{2 +}, \, \text{He}^{+}$ and $H$ is
NTA AbhyasNTA Abhyas 2020Atoms
Solution:
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 is the quantum number.
$⇒ \, \, \, \lambda \propto \frac{1}{\text{Z}^{2}}$
Atomic number of lithium is 3, of helium is 2 and of hydrogen is 1.
$\therefore \lambda_{\mathrm{Li}^{2+}}: \lambda_{\mathrm{He}^{+}}: \lambda_{\mathrm{H}}=\frac{1}{(3)^2}: \frac{1}{(2)^2}: 1$
$= \frac{1}{9} \text{:} \frac{1}{4} \text{:} 1$