InterviewSolution
Saved Bookmarks
| 1. |
A proton and an `alpha`-particle are accelerated through same potential difference. Find the ratio of their de-Brogile wavelength. |
|
Answer» `lambda=h/"mv"=h/sqrt(2mE)=h/sqrt(2mqV)[because E=qV]` For proton `m_p=m,q=e` For `alpha`-particle `m_alpha`=4 m, q=2 e `lambda_alpha/lambda_p=sqrt((m_pq_p)/(m_alphaq_alpha))` `=1/(2sqrt2)` |
|