Question

A particle (m = 3.0 × 10-28 kg) starting from rest, experiences an acceleration of 2.4 × 107 m/s2 for 5.0 s. What is its de Broglie wavelength λ at the end of this period?

Answer 1

The expression for the de-Broglie wavelength of the particle is, 2h р The expression for the momentum of the particle is, p=mv The expression of the kinematic equation for the particle is, Vo = 1; + at The initial velocity of the particle is zero, because, initially it is at rest. v; = 0 Then the final momentum of the particle is, p=mv. = m(v; +at) = m(0+ at = mat The particle has the wave nature, so, the particle may travel with speed of the light. The expression for the mass of the particle is, m=ymo The expression for the Lorentz constant is,

Write the expression for the de-Broglie wavelength of the particle as follows: ? mat m, at m, at The Planck's constant is, h=6.626x10-34 J.S The rest mass of the particle is, mo = 3.0x10-28 kg The acceleration of the particle is, a=2.4x10 m/s2 The time taken for the particle is, t=5.05

Calculate the de-Broglie wavelength of the particle as follows: :) h 2 = 4 hat [ (2.4x10' m/s2)(5.0s) 6.626x10-34 Jos ox10 J-s 3x108 m/s J ||(3.0x10-28 kg)(2.4x10’m/s)(5.0s) =1.686x10-14 m =16.86x10-15 m (or) 16.86 fm