A photon of wavelength $\lambda = 102.6\,\text{nm}$ is incident on a ground-state hydrogen atom. The atom can absorb it and transition to which excited state? (Take $hc = 1240\,\text{eV\,nm}$.)

1. Photon energy: $E = hc/\lambda = 1240/102.6 = 12.09\,\text{eV}$. 2. From the ground state $E_1 = -13.6\,\text{eV}$, the atom needs an energy gap of exactly $E_n - E_1$ for some integer $n \geq 2$. 3. Required final energy: $E_n = E_1 + 12.09 = -13.6 + 12.09 = -1.51\,\text{eV}$. 4. From $E_n = -13.6/n^2$: $-1.51 = -13.6/n^2 \Rightarrow n^2 = 13.6/1.51 = 9.00 \Rightarrow n = 3$. 5. So a $12.09\,\text{eV}$ photon resonantly drives the $n=1 \to n=3$ transition. This is the Ly-$\beta$ line. 6. Option A would require $E = 10.2\,\text{eV}$ (Ly-$\alpha$). Option C would require $E = 12.75\,\text{eV}$. Option D forgets that quantised levels admit specific photon energies — and $12.09$ is one of them. _Source: NCERT Class 12 Physics Part 2, Ch 12, §12.4.1 + §12.5 (energy level values and absorption), p. 10–11._