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

Explanation

We know that `lambda=(h)/sqrt(2mK)`
Since proton and electron have same de Broglie wavelength, we get
`(h)/sqrt(2m_(p) K_(p))=(h)/sqrt(2m_(e)K_(e))`
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))`
`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.