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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.