1. Use the hydrogen-like atom formula

For any one-electron species, the energy levels are $E_n = -13.6\,\frac{Z^2}{n^2}\text{ eV}$ where $Z$ is the atomic number.

For a transition from $n=2$ to $n=1$, $\Delta E = 13.6 Z^2\left(1-\frac14\right)=13.6 Z^2\cdot \frac34$ So, $\Delta E \propto Z^2$

Since emitted photon energy is $\Delta E = \frac{hc}{\lambda},$ we get $\lambda \propto \frac{1}{Z^2}$

2. Identify $Z$ for each species

• Hydrogen $\left({}_1H^1\right)$: $Z=1$
• Deuterium $\left({}_1H^2\right)$: $Z=1$
• Singly ionised helium $\left({}_2He^4\right)^+$: $Z=2$
• Doubly ionised lithium $\left({}_3Li^6\right)^{++}$: $Z=3$

Although hydrogen and deuterium have different nuclei, approximately their wavelengths are equal because both have the same nuclear charge $Z=1$. (A small reduced-mass correction is ignored here.)

Thus, $\lambda_1 \propto \frac{1}{1^2}=1$ $\lambda_2 \propto \frac{1}{1^2}=1$ $\lambda_3 \propto \frac{1}{2^2}=\frac14$ $\lambda_4 \propto \frac{1}{3^2}=\frac19$

3. Write the ratio

Therefore, $\lambda_1 = \lambda_2 = 4\lambda_3 = 9\lambda_4$

4. Match with the options

This matches Option C: $\lambda_1 = \lambda_2 = 4\lambda_3 = 9\lambda_4$

5. Comparison with stored correct answer

Stored correct answer: C

Our derived answer: C

So they agree.