An electron in the hydrogen atom initially in the fourth excited state makes a transition to $$n^{\text{th}}$$ energy state by emitting a photon of energy 2.86 eV. The integer value of n will be ________.

The energy of an electron in a hydrogen atom is given by
$$E_n = -\frac{13.6\text{ eV}}{n^{2}}$$

“Fourth excited state’’ means the electron starts from the level
$$n_i = 5$$  (because $$n = 1$$ is ground, $$n = 2$$ first excited, …).

If the electron finally reaches an unknown level $$n_f$$ and emits a photon of energy $$E_{\gamma}=2.86\text{ eV}$$, then by energy conservation

$$E_{\gamma}=E_{n_i}-E_{n_f}$$

Substituting the formula for $$E_n$$ gives

$$2.86 = \left(-\frac{13.6}{5^{2}}\right) - \left(-\frac{13.6}{n_f^{2}}\right)$$

Simplify the brackets:

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

Divide both sides by $$13.6$$ to isolate the bracketed term:

$$\frac{2.86}{13.6} = \frac{1}{n_f^{2}} - \frac{1}{25}$$

Calculate the numerical value:

$$\frac{2.86}{13.6} \approx 0.21$$

Add $$\frac{1}{25}=0.04$$ to both sides:

$$0.21 + 0.04 = \frac{1}{n_f^{2}}$$

$$\frac{1}{n_f^{2}} \approx 0.25$$

Taking reciprocal and square-root:

$$n_f^{2} \approx 4 \quad\Longrightarrow\quad n_f \approx 2$$

Since $$n_f$$ must be an integer quantum number, we take
$$n_f = 2$$.

Therefore, the electron drops to the second energy level. The required integer value of $$n$$ is 2.