A proton and an α-particle are accelerated through the same potential difference. Which one of the two has (i) greater de-Broglie wavelength, and (ii) less kinetic energy?

(i) The de-Broglie wavelength λ of a particle of mass m and charge e, accelerated by a potential difference V is given by λ=h√(2meV)
for proton λp=h√(2mpepV) eq1
for alpha λa=h√(2maeaV) eq2
dividing eq1 by eq2 we get λp/λa=√(maea/mpep)
now we know that ma=4mp, ea=2ep
therefore λp/λa=√(4mp2ep/mpep)=√8
or λp>λa
(ii) de-Broglie wavelength in terms of kinetic energy is given by λ=h√(2mK), or K=h²/2mλ²
for proton Kp=h²/2mpλp²
for alpha Ka=h²/2maλa²
or Kp/Ka=maλp²/mpλa²
or Kp/Ka=(4mpλp²)/(mp*8λa²)=1/2
or Kp<Ka