If the ionization energy of a hydrogen atom in its ground state is denoted as , the energy required — Atomic Structure Chemistry Question

💡 Solution & Explanation

The ionization energy $E_0 = 13.6 \text{ eV}$. The energy required to jump from $n=2$ to $n=3$ is $\Delta E = E_0 \times (\frac{1}{n_1^2} - \frac{1}{n_2^2}) = E_0 \times (\frac{1}{2^2} - \frac{1}{3^2}) = E_0 \times (\frac{1}{4} - \frac{1}{9}) = E_0 \times \frac{9-4}{36} = \frac{5}{36} E_0$. Thus, $y = 5$.