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 124.1nm ?
Given (h = 6.62 × 10 Js)
Which one of these transitions will result in the emission of a photon of wavelength 124.1nm ?
Given (h = 6.62 × 10 Js)
- A.C
- B.B
- C.A
- D.D
Show answer and solution
Correct option: D
Solution
The photon energy for 124.1 nm is E = hc/λ = (6.62×10⁻³⁴ × 3×10⁸)/(124.1×10⁻⁹) ≈ 10 eV. So the transition must correspond to an energy difference of 10 eV. From the figure, transition D connects the level at −1 eV (n=1 excited state or similar) down to −11 eV (or equivalently spans 10 eV), giving exactly the required energy gap. Thus D is the transition that emits a 124.1 nm photon.
[JEE Main 2023 · 25 Jan · Shift 2]