Kinetic energy of proton is the same as that of an `alpha`-particle. What is the ratio of their respective de Brogile wavelengths?

Explanation

Kinetic energy,
`E = (1)/(2) mv^(2) = ((mv)^(2))/(2m) or, mv = sqrt(2mE)`
Hence, de Broglie wavelength,
`lamda = (h)/(mv) = (h)/(sqrt(2mE))`
Since, kinetic energy of proton equals that of `alpha-`particle,
`(lamda_(1))/(lamda_(2)) = sqrt((m_(2))/(m_(1))) = sqrt((4)/(1)) = (2)/(1)`