A single electron orbits a stationary nucleus with atomic number . What will be the approximate ener — Atomic Structure Chemistry Question

💡 Solution & Explanation

The energy is given by $\Delta E = 13.6 \times Z^2 \times (\frac{1}{n_1^2} - \frac{1}{n_2^2}) \text{ eV}$. Here $Z=5, n_1=3, n_2=4$. So, $\Delta E = 13.6 \times 5^2 \times (\frac{1}{3^2} - \frac{1}{4^2}) = 13.6 \times 25 \times (\frac{1}{9} - \frac{1}{16}) = 340 \times \frac{7}{144} \approx 16.53 \text{ eV}$.