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?

Question

Alejandra Wu

Physics

9 January, 14:43

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?

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Answers (1)

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Kassidy Moyer · Answer 1

1.8 x 10⁻³⁴ m.

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.