1. Identify the energy levels involved

For hydrogen atom, $E_n=-\frac{13.6}{n^2}\,\text{eV}$

Also,

• ground state $\Rightarrow n=1$
• first excited state $\Rightarrow n=2$
• second excited state $\Rightarrow n=3$
• third excited state $\Rightarrow n=4$

2. First transition: second excited state to first excited state

This means $n=3 \to n=2$

The emitted photon energy is $\Delta E_1=13.6\left(\frac{1}{2^2}-\frac{1}{3^2}\right)$ $=13.6\left(\frac{1}{4}-\frac{1}{9}\right)$ $=13.6\left(\frac{5}{36}\right)$

If its wavelength is $\lambda_0$, then $\frac{hc}{\lambda_0}=13.6\cdot \frac{5}{36}$

3. Second transition: third excited state to second orbit

Third excited state means $n=4$, and second orbit means $n=2$. So transition is $n=4 \to n=2$

The emitted photon energy is $\Delta E_2=13.6\left(\frac{1}{2^2}-\frac{1}{4^2}\right)$ $=13.6\left(\frac{1}{4}-\frac{1}{16}\right)$ $=13.6\left(\frac{3}{16}\right)$

If its wavelength is $\lambda$, then $\frac{hc}{\lambda}=13.6\cdot \frac{3}{16}$

4. Use inverse proportionality of wavelength and energy

Since $\lambda \propto \frac{1}{\Delta E}$ we get $\frac{\lambda}{\lambda_0}=\frac{\Delta E_1}{\Delta E_2}$

So, $\frac{\lambda}{\lambda_0}=\frac{\frac{5}{36}}{\frac{3}{16}}=\frac{5}{36}\cdot\frac{16}{3}$ $=\frac{80}{108}=\frac{20}{27}$

Hence, $\lambda=\frac{20}{27}\lambda_0$

Given that $\lambda=\frac{20}{x}\lambda_0$ we get $x=27$

5. Final answer

$\boxed{27}$

The derived answer matches the stored correct answer.