de Broglie wavelength of the wave asociated with a moving electron and a proton are equal. Show the velocity of electron is greater than that of the proton.

Explanation

According to de-Broglie's theory application to moving microscopic particles like electron `lamda=(h)/(mv)` [m=mass of the moving particle, v=velocity of the moving partile].
Now if the mass and velocity of electron be `m_(e) and v_(e)` and the mass and velocity of proton be `m_(p) and v_(p)` respectively then according to the question.
`(h)/(m_(e)v_(e))=lamda=(h)/(m_(p)v_(p))" "therefore m_(e)v_(e)=m_(p)v_(p) or, (v_(e))/(v_(p))=(m_(p))/(m_(e))`
But, `m_(p) gt m_(e)` so, `v_(e) gt p_(p)` (proved)