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A proton and an electron have the same kinetic energy. Which one has a smaller de-Broglie wavelength and why?

Last updated date: 25th Jun 2024
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Hint: First according to the question kinetic energy of electron is equal to the kinetic energy of proton. Now we know the formula for de-Broglie wavelength. In the formula for de-Broglie wavelength, there are some fixed values we don’t need those value we need to omit those and find a proportionality relation for de-Broglie wavelength. Now we need to compare this relation to know the answer.

Formula used: $\lambda =\dfrac{h}{\sqrt{2mK}}$

Complete step by step answer:
According to the question we know that,
Kinetic energy of the electron = Kinetic energy of the proton.
Now we know that the formula for de-Broglie wavelength is,
$\lambda =\dfrac{h}{\sqrt{2mK}}$ ,
Now we can see that in the above formula the ‘h’, ‘K’ and ‘2’ are constant hence we can say that the de-Broglie wavelength is inversely proportional to ‘m’.
$\lambda =\dfrac{1}{\sqrt{m}}$,
Now, we know that the mass of electron is: $9.10\times \text{ }{{10}^{-31}}~$ Kg
And, the mass of the proton is: \[1.67\times {{10}^{-27}}\] Kg
So, we can clearly see that the mass of the proton is very much greater than the mass of the electron.
So, the de-Broglie wavelength of electron will be greater than the wavelength of proton, that is
${{m}_{e}}<{{m}_{p}}$ ,
${{\lambda }_{e}}>{{\lambda }_{p}}$.
So, in a summarized form, the mass of the electron is less so the de-Broglie wavelength is more because the kinetic energy is constant for both the particles.

Note: In the formula $\lambda =\dfrac{h}{\sqrt{2mK}}$, ‘h’ is the planck’s constant, and ‘K’ is the kinetic energy, we were already told that kinetic energy of both electron and proton are same so we are considering the kinetic energy also as a constant value. Students must remember the values of mass for some subatomic particle.