A proton and an electron have same velocity. Which one has greater de-Broglie wavelength and why?

Explanation

Since de-Broglie wavelenth `lambda` in term s of kinetic energy is given as
`lambda = (h)/(sqrt(2mk))`
Where `K` is kinetic energy, `m` is mass of electron and `h` is the Plank's constant.
Thus for electron and proton with same kinetic energy, de Broglie wavelength would depend on mass.
Since `lambda prop (1)/(sqrt(m))`
Now `m_(p) gt m_(e)` so, `lambda_(e) gt lambda_(p)`.
Hence wavelength of electron is greater than wavelength of proton.