MHT CET202120 Sep 2021Morning ShiftPhysicsDual Nature of MatterActual
The P.E. ' U ' of a moving particle of mass ' m ' varies with ' x ' is shown in the figure. The de-Broglie wavelength of the particle in the regions 0 x 1 and x>1 are ₁ and ₂ respectively. If the total energy of the particle is ' nE ', then the ratio ₁ / ₂ is
Options
- An^2 n-1
- Bn-1 n
- Cn n-1
- Dn(n-1) n
Correct answer
C. n n-1
Step-by-step solution
In the region 0 x 1 , the potential energy of the particle is E . Total energy is nE . Hence, kinetic energy, K = nE - E =( n -1) E Its momentum, p ₁= 2 mK = 2 ~m ( n -1) E ₁= h p ₁ = h 2 ~m ( n -1) E In the region x >1, PE is zero Hence, total is kinetic energy. aligned & K = nE & ₂= h p ₂ = h 2 mnE & ₁ ₂ = n n -1 aligned