Saved Bookmarks
| 1. |
Ligth of wavelength `4000 Å` is incident on a metal plate whose work function is `2 eV`. What is maximum kinetic enegy of emitted photoelectron ?A. `0.5 eV`B. `1.1 eV`C. `2.0 eV`D. `1.5 eV` |
|
Answer» Correct Answer - B (b) If the maximum kinetic energy of photo electrons emitted from metal surface is `E_(k)` and `W` is the work-function of metal then `E_(k) = (hc)/(lambda) - W` where `hv` is the energy of photon absorbed by the electron in metal. `: E_(k) = (hc)/(lambda) - W` where `v = (c )/(lambda)` Putting the numerical values, we have `E_(k) = [(6.6 xx 10^(-34) xx 3 xx 10^(8))/(4000 xx 10^(-10) xx 1.6 xx 10^(-19)) - 2] eV` `E_(k) = 3.1 - 2 = 1.1 eV` |
|