1. Identify the energy levels involved

In hydrogen atom:

• First excited state $\Rightarrow n=2$
• Second excited state $\Rightarrow n=3$
• Third excited state $\Rightarrow n=4$

So the transitions are:

• First: $n=4 \to n=3$ with wavelength $\lambda_1$
• Second: $n=3 \to n=2$ with wavelength $\lambda_2$

2. Use the hydrogen energy difference relation

For hydrogen, the emitted photon energy is $E = 13.6\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right) \text{ eV}$

Also, $E=\frac{hc}{\lambda} \Rightarrow \lambda \propto \frac{1}{E}$

Thus, $\frac{\lambda_1}{\lambda_2} = \frac{E_2}{E_1}$ where $E_1$ corresponds to $4\to 3$ and $E_2$ corresponds to $3\to 2$.

3. Calculate energy for $4\to 3$

$E_1=13.6\left(\frac{1}{3^2}-\frac{1}{4^2}\right)$ $=13.6\left(\frac{1}{9}-\frac{1}{16}\right)$ $=13.6\left(\frac{16-9}{144}\right)$ $=13.6\cdot \frac{7}{144}$

4. Calculate energy for $3\to 2$

$E_2=13.6\left(\frac{1}{2^2}-\frac{1}{3^2}\right)$ $=13.6\left(\frac{1}{4}-\frac{1}{9}\right)$ $=13.6\left(\frac{9-4}{36}\right)$ $=13.6\cdot \frac{5}{36}$

5. Find the ratio of wavelengths

$\frac{\lambda_1}{\lambda_2} = \frac{E_2}{E_1}$ $=\frac{13.6\cdot \frac{5}{36}}{13.6\cdot \frac{7}{144}}$ $=\frac{5}{36}\cdot \frac{144}{7}$ $=\frac{5\cdot 4}{7}$ $=\frac{20}{7}$

6. Match with options

$\frac{\lambda_1}{\lambda_2}=\frac{20}{7}$

So the correct option is D.