JEE Main 2021 · 31 August · evening · Single correct · +4 / −1
A free electron of 2.6 eV energy collides with a H⁺ ion. This results in the formation of a hydrogen atom in the first excited state and a photon is released. Find the frequency of the emitted photon. (h = 6.6 × 10⁻³⁴ Js)
- (a) 1.45 × 10¹⁶ MHz
- (b) 0.19 × 10¹⁵ MHz
- (c) 1.45 × 10⁹ MHz
- (d) 9.0 × 10²⁷ MHz
Answer and solution
Answer (c) 1.45 × 10⁹ MHz
Solution
Physics sir · 4 steps
The photon carries away the electron's kinetic energy plus the 3.4 eV it loses by getting bound in n = 2.
- Energy before. A free electron far from the proton has total energy equal to its kinetic energy, +2.6 eV.
- Energy after. In hydrogen, Eₙ = -13.6/n² eV. The first excited state is n = 2, so E₂ = -13.6/4 = -3.4 eV.
hν = 2.6 - (-3.4) = 6.0 eV = 6.0 × 1.6 × 10⁻¹⁹ J = 9.6 × 10⁻¹⁹ J
- Find frequency. ν = (9.6 × 10⁻¹⁹)/(6.6 × 10⁻³⁴) = 1.45 × 10¹⁵ Hz. Since 1 MHz = 10⁶ Hz, this is 1.45 × 10⁹ MHz.
- So. The emitted photon has frequency 1.45 × 10⁹ MHz, which is option (c).
The idea to keep. When a free electron is captured into level n, the photon energy is its kinetic energy plus the binding energy of that level, 13.6/n² eV. Convert eV to joules before dividing by h.
Answer from the NTA final answer key. Solution by PrepWiser.