When the electron in a hydrogen atom makes a transition from $n = 2$ (first excited state) DIRECTLY back to $n = 1$ (ground state), the energy of the EMITTED photon is

1. Bohr's third postulate: when an electron drops from a higher energy $E_i$ to a lower energy $E_f$, a photon of energy $h\nu = E_i - E_f$ is emitted. 2. Use $E_n = -13.6/n^2\,\text{eV}$. 3. Initial: $E_2 = -13.6/4 = -3.40\,\text{eV}$. 4. Final: $E_1 = -13.6/1 = -13.6\,\text{eV}$. 5. Photon energy: $h\nu = E_2 - E_1 = -3.40 - (-13.6) = 10.20\,\text{eV}$. 6. This is the first Lyman line (often denoted $\text{Ly-}\alpha$). Option A is just $|E_2|$. Option C is $E_3 - E_1 = 12.09\,\text{eV}$ (Ly-β). Option D is the ionisation energy itself. _Source: NCERT Class 12 Physics Part 2, Ch 12, §12.4.1 (Eq. 12.10 + numerical values), p. 10._