The ratio of the wavelengths for 2 → 1 transition in Li++, He+ and H is :
Answer & explanation
Correct answer: option 3
$\frac{1}{\lambda}=R Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \Rightarrow \lambda \propto \frac{1}{Z^2}$
$\lambda_{L^{++}}: \lambda_{H e^{+}}: \lambda_H$ = 4 : 9 : 36