JEE Main 2022 · 27 July · morning · Single correct · +4 / −1
An electron (mass m) with an initial velocity v⃗=v₀ î(v₀>0) is moving in an electric field E⃗=-E₀ î(E₀>0) where E₀ is constant. If at t=0 de Broglie wavelength is λ₀=h/(m v₀), then its de Broglie wavelength after time t is given by
- (a) λ₀
- (b) λ₀(1+(e E₀ t)/(m v₀))
- (c) λ₀ t
- (d) λ₀/(1+(e E₀ t)/(m v₀))
Answer and solution
Answer (d) λ₀/(1+(e E₀ t)/(m v₀))
Solution
Physics sir · 3 steps
Field points along minus x, so it pushes the negative electron along plus x: it speeds up and λ shrinks.
- Direction of force. The force on the electron is F⃗ = -eE⃗ = -e(-E₀î) = eE₀î. It points along +x, the same way the electron is already moving.
- Velocity after t. The acceleration is constant, a = eE₀/m, so v = v₀ + eE₀/mt = v₀(1 + (eE₀ t)/(m v₀)).
λ = h/mv = h/(m v₀(1+(eE₀ t)/(m v₀))) = λ₀/(1+(eE₀ t)/(m v₀))
- So. The electron speeds up, so its momentum grows and its wavelength falls. The answer is (d).
The idea to keep. First find the force with F⃗ = qE⃗ and the sign of the charge. Then remember λ = h/p: if momentum goes up, wavelength goes down in the same ratio.
Answer from the NTA final answer key. Solution by PrepWiser.