A photon is emitted in transition from $$n = 4$$ to $$n = 1$$ level in hydrogen atom. The corresponding wavelength for this transition is (given, $$h = 4 \times 10^{-15}$$ eV s)

$$\Delta E = E_{n_i} - E_{n_f} = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \text{ eV}$$

$$\Delta E = \frac{hc}{\lambda} \implies \lambda = \frac{hc}{\Delta E}$$

$$\Delta E = 13.6 \left( \frac{1}{1^2} - \frac{1}{4^2} \right)$$

$$\Delta E = 13.6 \left( \frac{15}{16} \right) = 12.75 \text{ eV}$$

$$hc = (4 \times 10^{-15} \text{ eV}\cdot\text{s}) \times (3 \times 10^8 \text{ m/s})$$

$$hc = 12 \times 10^{-7} \text{ eV}\cdot\text{m}$$

$$hc = 12 \times 10^{-7} \times 10^9 \text{ eV}\cdot\text{nm} = 1200 \text{ eV}\cdot\text{nm}$$

$$\lambda = \frac{1200 \text{ eV}\cdot\text{nm}}{12.75 \text{ eV}}$$

$$\lambda \approx 94.117 \text{ nm}$$