1. Identify the initial state

The fourth excited state means:

• Ground state $= 1$
• First excited state $= 2$
• Second excited state $= 3$
• Third excited state $= 4$
• Fourth excited state $= 5$

So the electron is initially in: $n_i = 5$

2. Use the hydrogen energy level formula

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

If the electron falls from $n_i=5$ to $n$, the emitted photon energy is: $\Delta E = 13.6\left(\frac{1}{n^2} - \frac{1}{5^2}\right)\text{ eV}$

Given: $\Delta E = 2.86\,\text{eV}$

So, $13.6\left(\frac{1}{n^2} - \frac{1}{25}\right)=2.86$

3. Solve for $n$

Divide both sides by $13.6$: $\frac{1}{n^2} - \frac{1}{25} = \frac{2.86}{13.6}$

Now, $\frac{2.86}{13.6} \approx 0.2103$

Thus, $\frac{1}{n^2} = 0.2103 + 0.04 = 0.2503 \approx 0.25$

Hence, $\frac{1}{n^2} = \frac{1}{4}$

So, $n^2 = 4 \Rightarrow n=2$

4. Check

Energy of level 5: $E_5 = -\frac{13.6}{25} = -0.544\,\text{eV}$

Energy of level 2: $E_2 = -\frac{13.6}{4} = -3.4\,\text{eV}$

Photon energy emitted: $|E_2 - E_5| = 3.4 - 0.544 = 2.856\,\text{eV} \approx 2.86\,\text{eV}$

So the result is correct.