Saved Bookmarks
| 1. |
For given metal work function is W_(0) and threshold wavelength is lambda_(0).If metal has work function (W_(0))/(2) then what will be threshold wavelength ? |
|
Answer» `(lambda_(0))/(4)` `THEREFORE (W_(0))/(W_(0))=(lambda_(0))/(lambda_(0))` `(W_(0))/(W_(0))=(lambda_(0))/(lambda_(0))` `(W_(0))/((W_(0))//(2))=(lambda_(0))/(lambda_(0)) [because W_(0)=(W_(0))/(2)]` `therefore 2lambda_(0)=lambda_(0)` |
|