Question:

An α-particle and a proton are accelerated from rest by a potential difference of 100V. After this, their de Broglie wavelengths are λα and λp respectively. The ratio λp/λα, to the nearest integer, is:

We have, 1/2mv² = eV
De broglie wavelength = λ = h/mv
λ = h/√(2meV)
λp = h/√[2(m)eV] (1)
λα = h/√[2(4m)eV] (2)
Now, λp/λα = √8 ≈ 3