Saved Bookmarks
| 1. |
Find the frequency of light which ejects electrons from a metal surface stopped by a retarding potential of 3.3 V. if photoelectric emission begins in this metal at a frequency of 8xx10^(14)Hz, calculate the work function (in eV) for this metal. |
|
Answer» SOLUTION :Here stopping potential `V_(0)=3.3V` and threshold frequency `v_(0)=8xx10^(14)Hz` `THEREFORE` Work function `phi_(0)=(hv_(0))/(e)eV=(6.63xx10^(-34)xx8xx10^(14))/(1.6xx10^(-19))eV=3.315eV` If frequency of INCIDENT light be V, then `(hv)/(e)=[phi_(0)+V_(0)]eV` `implies v=[(phi_(0)+V_(0))/(h)]*e=((3.315xx3.3)xx1.6xx10^(-19))/(6.63xx10^(-34))=1.58xx10^(15)Hz`. |
|