A proton and a deuteron are accelerated through the same accelerating potential. Which one of the two has a greater value of de-Broglie wavelength associated with it, and less momentum?

From the law of conservation of energy, Kinetic energy equals the change in electrical energy in the particle.
1/2mv² = qV
p = mv = √(2mqV)
De-broglie wavelength of a particle is given by:
λ = h/p
λ = h/√(2mqV)
(a) Ratio of de-broglie wavelength of proton to deuteron is given by:
λp/λd = √(md/mp)
Since md (mass of deuteron) = 2mp (mass of proton)
λp/λd = √(2mp/mp) = √2
Hence, the proton has a higher de-broglie wavelength.
(b) p = h/λ
Hence, the deuteron has higher momentum.