JEE Main20204 Sep 2020Morning ShiftPhysicsDual Nature of MatterActual
Particle A of mass m A = m 2 moving along the x -axis with velocity v 0 collides elastically with another particle B at rest having mass m B = m 3 . If both the particles move along the x -axis after the collision, the change ∆ λ in the wavelength of the particle A , in terms of its de-Broglie wavelength λ 0 before the collision is:
Options
- A∆ λ = 3 2 λ 0
- B∆ λ = 5 2 λ 0
- C∆ λ = 2 λ 0
- D∆ λ = 4 λ 0
Correct answer
D. ∆ λ = 4 λ 0
Step-by-step solution
v 1 = 2 m / 3 0 m 2 + m 3 + m 2 − m 3 v m 2 + m 3 = v 5 For particle A , Initial de-Broglie wavelength λ 0 = h m 2 v = 2 h mv Final de-Broglie wavelength after collision. λ 1 = h m 2 v 5 = 10 h mv = 5 λ 0 Change in De-Broglie wavelength Δλ = λ i − λ 0 = 4 λ 0