The energy levels of an atom is shown in figure.
Which one of these transitions will result in the emission of a photon of wavelength 124.1 nm?
Given (h = 6.62 × 10⁻³⁴ J·s)

Solution

Physics sir · 2 steps

Turn 124.1 nm into electron-volts first, then look for the arrow that spans exactly that energy gap.

1. Photon energy. The emitted photon carries the energy difference between the two levels, E = hc/λ. Take c = 3 × 10⁸ m s⁻¹ and 1 eV = 1.6 × 10⁻¹⁹ J.

   E = (6.62 × 10⁻³⁴ × 3 × 10⁸)/(124.1 × 10⁻⁹) = 1.6 × 10⁻¹⁸ J = 10.0 eV

   Transition	From → to	Energy gap	Wavelength
   A	0 → -2.2 eV	2.2 eV	about 564 nm
   B	0 → -5.2 eV	5.2 eV	about 239 nm
   C	-2.2 → -5.2 eV	3.0 eV	about 413 nm
   D	0 → -10.0 eV	10.0 eV	124.1 nm

2. So. Only transition D releases 10.0 eV, the energy of a 124.1 nm photon. The answer is (d).

The idea to keep. Remember hc ≈ 1240 eV nm, so E(eV) = 1240/λ(nm). Then just match that energy to the gap between two levels.

Answer from the NTA final answer key. Solution by PrepWiser.