99 Percentile Qs Bank for JEE MainPhysicsDual Nature of Matter
The potential energy of a particle varies as U x = E 0 for 0 ≤ x ≤ 1 = 0 for x > 1 For 0 ≤ x ≤ 1 , the de-Broglie wavelength is λ 1 and for x > 1 the de-Broglie wavelength is λ 2 . The total energy of the particle is 2 E 0 . Find λ 1 λ 2
Options
- A3
- B7
- C2
- D5
Correct answer
C. 2
Step-by-step solution
For 0 ≤ x ≤ 1 , U = E 0 ∴ Kinetic energy, K 1 = Total energy - U = 2 E 0 - E 0 = E 0 ∴ λ 1 = h 2 m E 0 For x > 1 , U = 0 ∴ Kinetic energy K 2 = Total energy - U = 2 E 0 ∴ λ 2 = h 2 m ( 2 E 0 ) ...(ii) From Eqs. (i) and (ii), we have λ 1 λ 2 = 2