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Question: What is the de Broglie wavelength of a heavier particle?...

What is the de Broglie wavelength of a heavier particle?

Explanation

Solution

In order to solve this question, we are going to firstly give the relation for the de Broglie wavelength and then discuss its dependence on mass. We will discuss the cases for the lighter and the heavier particles and what is the relation between their de Broglie wavelengths as calculated.

Formula used: The de- Broglie wavelength is given by:
λ=hmv\lambda = \dfrac{h}{{mv}}
Where, hhis Planck's constant, mmis the mass of the particle, vvis the velocity of the particle.

Complete step-by-step solution:
The de- Broglie wavelength is given by:
λ=hmv\lambda = \dfrac{h}{{mv}}
Where, hhis Planck's constant, mmis the mass of the particle, vvis the velocity of the particle.
Thus, the de-Broglie’s wavelength depends on the two factors, mass and the velocity of the particle. If the two particles are moving with the same velocity, then, the factor of mass comes into action. In the formula, we can see that the wavelength of a particle is inversely proportional to the mass of the particle. That is, the more is the mass of the particle, the less is the wavelength and vice versa. Thus, the de-Broglie wavelength of a heavier particle will be smaller as compared to that of a lighter particle.

Note: Alternatively, If we consider two masses, mmandMMsuch that m<Mm < M
Then the de Broglie wavelengths are as follows:
λ=hmv\lambda = \dfrac{h}{{mv}}andλ=hMv\lambda = \dfrac{h}{{Mv}}
Thus, hmv>hMv\dfrac{h}{{mv}} > \dfrac{h}{{Mv}}
This means that the de Broglie wavelength for the particle of massMM, i.e., the heavier particle is smaller as compared to the de Broglie wavelength for the particle of massmm, i.e., the lighter particle.