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?

Answer 1

As `" "lamda=h/sqrt(2mqv)`
`therefore" "lamdaprop1/(sqrt(mq))`
`" "(lamda_(p))/(lamdaalpha)=(sqrt(m_(alpha)q_(alpha)))/(sqrt(m_(p)q_(p)))=(sqrt(4m_(p)xx2e))/(sqrt(m_(p)xxe))=sqrt8`
`therefore" "lamda_(p)=sqrt8lamdaalpha`
i.e., wavelength of proton is `sqrt8` times wavelength of `alpha`-particle.