Saved Bookmarks
| 1. |
Derive the expression for a due Broglie wavelength lambda of a relativistic particle moving with kinetic energy T. At what value of T does the error in deremining lambda using the non-relativistic forumula not exceed 1% for an electron and a proton? |
|
Answer» Solution :For a relative particle `T=mc^(2)= ` total energy `=sqrt(c^(2)p^(2)+m^(2)c^(4))` squaring `sqrt(T(T+2MC^(2)))=CP` Hence `lambda=(2pi ħc)/(sqrt(T(T+2mc^(2))))` `=(2pi ħ)/(sqrt(2mT(1+(T)/(2mc^(2)))))` If we use non relativistic FORMULA, `lambda_(NR)=(2pi ħ)/(sqrt(2mT))` `(Delta lambda)/(lambda)=(lambda_(NR)-lambda)/(lambda_(NR))~=(T)/(4MC^(2))` (If `T//2mc^(2)lt lt1`, we can write`(1+(T)/(2mc^(2)))^(1//2)~= 1-(T)/(4mc^(2)))` Thus `T le (4mc^(2)Delta lambda)/(lambda)` if the ERROR is less than `Delta lambda` For electron the error is not more than `1%` if ` T le 4xx0.511xx0.1MeV` For a proton, the error is not more than `1%` if `T le 4xx938xx0.01MeV` i.e., `T le 37.5MeV`. |
|