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 λpλα, to the nearest integer, is:
We have, 12mv2=eV
De broglie wavelength= λ=hmv
λ=h√2meV
λp=h√2(m)eV....(1)
λα=h2(4m)2eV....(2)
Now, λpλα=√8≃3