Question

Deduce the condition when the De-Broglie wavelength associated with an electron would be equal to that associated with a proton if a proton is `1836` times heavier than an electron.

Answer 1

Correct Answer - `v_(e)=1836 v_(p)`
Debroglie wavelength associated with particle of mass `(m)` moving with velocity `(v)` is
`lambda=h/(mv) rArr lambda_(P)=h/(m_(P)v_(P))` and `lambda_(e)=h/(m_(e)v_(e))`
given `lambda_(p)=lambda_(e) rArr (h)/(m_(p)v_(p))=h/(m_(e)v_(e)) rArr m_(p)v_(p)=m_(e)v_(e)`
`v_(e)/v_(p)=m_(p)/m_(e)=1836 rArr v_(e)=1836 v_(p)`
It means when velocity of electron will be `1836` times velocity of proton, then debroglie wavelength associated with electron would be equal to debroglie wavelength associated with proton.