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Question

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.


A

λe= λp

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B

λe< λp

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C

λe> λp

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D

The relation between λe and λpdepends on the accelerating potential difference

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Solution

The correct option is C

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 = h2meqVλp=h2mpqV

q and V is same for both

only one variable

me<mp

λe>λp

Option c is correct


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