Saved Bookmarks
| 1. |
The stopping potential for photoelectrons emitted from a surface illuminated by light wavelength of `5893A` is `036 V`. Calculate the maximum kinetic energy of photoelectrons, the work function of the surface, and the threshold frequency. |
|
Answer» We know that `KE_(max)=hf-phi((hc)/(lamda))-phi` or `phi=(hc)/(lamda)-KE_(max)` Also, `KE_(max)=ev_s=0.36eV` `impliesphi=((6.62xx10^(34))xx(3xx10^(8)))/(5893xx10^(-10))-0.36xx1.6xx10^(-19)` `=1.746eV` The threshold frequency is given by `f_0=(phi)/(h)=(2.794xx10^(-19))/(6.62xx10^(-34))=4.22xx10^(14)Hz` |
|