Question

A proton and an `alpha`-particle are accelerated, using the same potential difference. How are the de-Broglie wavelengths `lambda_(p) and lambda_(a)` related to each other?
A. `sqrt2`
B. `sqrt3`
C. `sqrt8`
D. `sqrt10`

Answer 1

Correct Answer - C
As,`lambda=h/sqrt(2mqv),thereforelambdaprop1/sqrt(mq)`
`lambda_(p)/lambda_(alpha)=sqrt(m_(alpha)q_(alpha))/sqrt(m_(p)q_(q))=sqrt(4m_(p)xx2e)/sqrt(m_(p)xxe)=sqrt8`
`therefore lambda_(p)=sqrt8lambda_(alpha)`
i.e., wavelength of proton is `sqrt8` times wavelength of `alpha`-particle.