Question

Particle A is at rest, and particle B collides head-on with it. The collision is completely inelastic, so the two particles stick together after the collision and move off with a common velocity. The masses of the particles are different, and no external forces act on them. The de Broglie wavelength of particle B before the collision is 1.8 × 10^-34 m. What is the de Broglie wavelength of the object that moves off after the collision?

Answer 1

1.8 x 10⁻³⁴ m.

Step-by-step explanation:

de Broglie wavelength ( λ ) of a moving particle is given by the following expression

λ = h / momentum of the particle

In other words, de Broglie wavelength depends upon the momentum of the particle.

In the given case , particle A which is stationary collides with another particle B having some momentum . After the collision , they move together.No external force acts on them . Therefore after the collision , their momentum will be conserved. In other words , their momentum remains the same as earlier. So their de Broglie wave length will also be the same as earlier , since it depends on the momentum of the moving body.

Hence the de Broglie wavelength of the object will be 1.8 x 10⁻³⁴ m.