JEE Main 2025 · 7 April · evening · Single correct · +4 / −1
A photoemissive substance is illuminated with a radiation of wavelength λᵢ so that it releases electrons with de-Broglie wavelength λₑ. The longest wavelength of radiation that can emit photoelectron is λₒ. Expression for de-Broglie wavelength is given by:
(m: mass of the electron, h: Planck's constant and c: speed of light)
- (a) λₑ = √(h λᵢ)/√2mc
- (b) λₑ = h/√(2mc ( 1/λᵢ - 1/λₒ ))
- (c) λₑ=√(h/(2 mc(1/λᵢ-1/λₒ)))
- (d) λₑ=√((h λ₀)/(2 m c))
Answer and solution
Answer (c) λₑ=√(h/(2 mc(1/λᵢ-1/λₒ)))
Solution
Physics sir · 3 steps
Get the kinetic energy from Einstein's equation, turn it into momentum, then use λ = h/p.
- Kinetic energy. The work function is the energy of a photon at the threshold wavelength, φ = hc/λₒ. Einstein's equation then gives K = hc/λᵢ - hc/λₒ = hc(1/λᵢ - 1/λₒ).
- Momentum. For the electron, p = √2mK = √2mhc(1/λᵢ - 1/λₒ).
λₑ = h/p = h/√2mhc(1/λᵢ-1/λₒ) = √(h/2mc(1/λᵢ-1/λₒ))
- So. Writing the outside h as √h² cancels one h under the root. That leaves option (c).
The idea to keep. For photoelectrons, always go energy, then momentum, then wavelength: K = hc(1/λᵢ-1/λₒ), p = √2mK, λ = h/p. Simplify the powers of h carefully at the end.
Answer from the NTA final answer key. Solution by PrepWiser.