Saved Bookmarks
| 1. |
X-rays fall on a photosensitive surface to cause photoelectric emission. Assuming that the work function of the surface can be neglected, find the relation between the de-Broglie wavelength (lambda) of the electrons emitted to the energy (E_(v)) of the incident photons. Draw the nature of the graph for lambda as a function of E_(v). |
|
Answer» <P> Solution :`E_(V)=phi_(0)+K_("MAX")`As`phi_(0)=0` `implies E_(v)=K_("max")` `implies K_("max")=(p^(2))/(2m)=E_(v)` `implies p=sqrt(2mE_(v))` `:.` Wavelength `(lambda)` of emitted electrons , `lambda=(h)/(p)=(h)/(sqrt(2mE_(v)))` 
|
|