Question

The potential energy of a partical varies as . U(x) = E_0 for 0 le x le 1 = 0 for x gt 1 For 0 le x le 1, de- Broglie wavelength is lambda_1 and for xgt the de-Broglie wavelength is lambda_2. Total energy of the partical is 2E_0. find (lambda_1)/(lambda_2).

Answer 1

λ₁/λ₂ = √2

Explanation:

de-Broglie wavelength of the particle is given by the formula:

λ = h/√(2mK)

Where,

K = Kinetic energy of a particle

For 0 ≤ x ≤ 1, U(x) = E₀

Total energy = Kinetic energy + Potential energy

2E₀ = K₁ + E₀

K₁ = 2E₀ - E₀

∴ K₁ = E₀

Now, the wavelength is given as:

λ₁ = h/√(2mK₁)

∴ λ₁ = h/√(2mE₀) → (equation 1)

For x > 1, U(x) = 0

∴ K₂ = 2E₀

λ₂ = h/√(2mK₂)

∴ λ₂ = h/√(2m(2E₀)) → (equation 2)

On dividing equation (1) by (2), we get,

λ₁/λ₂ = (h/√(2mE₀))/(h/√(2m(2E₀)))

λ₁/λ₂ = h/√(2mE₀) × √(2m(2E₀))/h

∴ λ₁/λ₂ = √2