Saved Bookmarks
| 1. |
Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate the threshold frequency and work function (W_(0)) of the metal. |
|
Answer» Solution :Threshold wavelength `(lambfa_(0))=6800 Å=6800xx10^(-10) m` As `c=V lambda :. v_(0)=c/lambda_(0)=(3.0xx10^(8)" ms"^(-1))/(6800xx10^(-10) m)=4.41xx10^(14) s^(-1)` Work FUNCTION `(W_(0))=hv_(0)=(6.626xx10^(-34) J s)(4.41xx10^(14) s^(-1))=2.92xx10^(-19) J` |
|