An electron (mass m) with initial velocity v⃗ = v₀î + v₀ĵ is in an electric field E⃗ = - E₀k̂. If λ ₀ is initial de-Broglie wavelength of electron, its de-Broglie wave length at time t is given by :

Solution

Physics sir · 4 steps

The field pushes the electron along z, so its speed grows. Track the speed, and λ = h/mv follows.

1. Force on electron. The electron's charge is -e, so the force is -eE⃗ = +eE₀ k̂. It accelerates along +z with a = eE₀/m, and the x and y velocities stay v₀.
2. Speed at time t. The velocity becomes v₀î + v₀ĵ + (eE₀ t)/mk̂, so v² = 2v₀² + e²E₀²t²/m². The initial speed is √2 v₀, not v₀.
3. Take the ratio. Since λ = h/mv, the wavelength is inversely proportional to the speed. Divide the new wavelength by λ₀ = h/(m√2 v₀).

   λ/λ₀ = (√2 v₀)/√(2v₀² + e²E₀²t²/m²) = 1/√(1 + e²E₀²t²/2m²v₀²)

4. So. λ = λ₀/√(1 + e²E₀²t²/2m²v₀²), which is option (c).

The idea to keep. For a changing de Broglie wavelength, find the full speed vector first. Then λ ∝ 1/v, and taking the ratio with the initial speed cancels h and m.

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