A proton and an `alpha`particle are accelerated through the same potential difference. Which one of the two has
greater de Broglie wavelength,

Explanation

Kinetic energy of the proton is 1 eV and that of alpha particle is 2 eV.
Thus,
`(mv_p^2)/2=1eV" "....(i)`
`((4m)v_2^2)/2=2eV" "...(ii)`
from (i) and (ii) , we get
`v_p=sqrt2v_a`
Thus, velocity of alpha particle is less than that of proton.
For de Broglie wavelength,
`lamda=h/(mv)`
For proton ,
`lamda_p=h/(mv_p)`
`implieslamda_p=h/(msqrt2v_a)`
For alpha, particle,
`lamda_a=h/(4mv_a)`
`implieslamda_p=2sqrt2lamda_a`
Thus de Broglie wavelength of proton will be greater than of alpha particle.