A proton and an electron are accelerated by the same potential difference. Letλe and λp enote the de Broglie wavelengths of the electron and the proton respectively.
λe> λp
Given – a proton and an electron are accelerated by same potential difference Let that be v FineTheir de Broglie wavelength is λe and λp
λe=hpeλp=hppλe=hmeveλp=hmpvP
So as we see here that the wavelength depends on 2 variables.
We cant compare as such.
Asme<mp yes
But we don’t know the relation betweenVeandVp
Or do we
There is an extra piece of information
The V-potential difference for accelerating both particles is same
So ………………
So if both had 0 velocity initially, then they were accelerated by same potential to gain velocity VeandVp
Lets find a relation
Lets conserve energy
Δ K.E.=12mv2e
Δ P.E.=qv
⇒12mev2e=qv
12mev2e=meqv
P=meve=√2meqv
similarlyP=meve=√2mpqv
Substituting in de Broglie’s wavelength
λe = h√2meqV∣λp=h√2mpqV
q and V is same for both
only one variable
me<mp
⇒ λe>λp
Option c is correct