Saved Bookmarks
| 1. |
Electrons with de-Broglie wavelength `lambda` fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-ray isA. `lamda_0 = ( 2mc lamda ^(2))/( h )`B. ` lamda_0 = ( 2h ) /( mc )`C. `lamda _0 = ( 2 m^(2) c ^(2) lamda ^(3))/(h^(2)) `D. `lamda _0 = lamda ` |
| Answer» Correct Answer - A | |