CBSE Class 12 · 2009 · Modern PhysicsShort Answer

The photon energy for wavelength $275\text{nm}$ is

(a) Computing $\Delta E$ for the marked transitions (levels at $0,-2,-4.5,-10\text{eV}$): the transition from the $0$ eV level to the $-4.5$ eV level gives $\Delta E=4.5\text{eV}$ ($\lambda \approx 275\text{nm}$) — this is transition B.
(b) Maximum wavelength corresponds to minimum energy. The smallest gap, $0\to -2\text{eV}$ ($\Delta E=2\text{eV}$, $\lambda \approx 620\text{nm}$), is transition A, which gives the maximum wavelength.