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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

  1. An^2 n-1
  2. Bn-1 n
  3. Cn n-1
  4. 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

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