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An electron, a proton and an alpha particle have KE of 16E, 4E and E respectively. What is the qualitative order of their de-Broglie wavelength?

A. λe>λp=λα
B. λp=λα>λe
C. λp>λe>λα
D. λα=λe≫λp

User Remon
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Final answer:

The qualitative order of the de Broglie wavelength for the electron, proton, and alpha particle is λe>λp=λα.

Step-by-step explanation:

The de Broglie wavelength of a particle is given by the formula λ = h/p, where λ is the wavelength, h is the Planck's constant, and p is the momentum of the particle. The momentum of a particle can be calculated using the formula p = √(2mKE), where m is the mass of the particle and KE is the kinetic energy.

In the given question, the electron has the highest kinetic energy, followed by the proton and the alpha particle. Since the de Broglie wavelength is inversely proportional to the momentum, we can infer that the electron will have the shortest de Broglie wavelength, followed by the proton and the alpha particle. Therefore, the qualitative order of their de Broglie wavelength is A. λe>λp=λα.

User JNDPNT
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