A proton and an electron have same de Broglie wavelength. Which of them moves faster and which possesses more kinetic energy ?

Explanation

We know that `lambda = (h)/(sqrt(2 mK))`
Since proton and electron have same de Broglie wavelength, we get
`(h)/(sqrt(2m_(p)K_(p))) = (h)/(sqrt(2m_(e) K_(e))) (or) (K_(p))/(K_(e)) = (m_(e))/(m_(p))`
Since `m_(e) lt m_(p), K_(p) lt K_(e)`, the electron has more kinetic energy than the proton.
`(K_(p))/(K_(e)) = ((1)/(2) m_(p)v_(p)^(2))/((1)/(2) m_(e)v_(e)^(2)) (or) (v_(p))/(v_(e)) = sqrt((k_(p)m_(e))/(k_(e)m_(p)))`
`(v_(p))/(v_(e)) = sqrt((m_(e)^(2))/(m_(p)^(2))) = (m_(e))/(m_(p))` since `(K_(p))/(K_(e)) = (m_(e))/(m_(p))`
Since `m_(e) lt m_(p), v_(p) lt v_(e)`, the electron moves faster than the proton.