Saved Bookmarks
| 1. |
A monochromatic light source of frequency villuminates a metallic surface and ejects photoelectron. The electron with maximum energy are just able to ionize the hydrogen atom in the ground state. When the whole experiment is repeated with an incident radiation of frequency 5/6 v, the photoelectrons so emitted are able to excite the hydrogen atom beam which then emits a radiation of the wavelength 1215Å. Find the work function of the metal : |
|
Answer» 6.875 eV `hv=W+(1)/(2)mv_(1)^(2)=w+` Ionization energy Now, ionisation potential =13.6V `:.hv=w+1.6xx10^(-19)xx13.6` `=w21.+76xx10^(-19) ""...(1)` In second case `h.(5)/(6)=w+(1)/(2) mv_(2)^(2)=w+(HC)/(lambda)` `=w+(6.6xx10^(-34)xx3xx10^(8))/(1215xx10^(-10))` `=w+16.3xx10^(-19) ""....(2)` Dividing (1) by (2), `(6)/(5)=(w+21.76xx10^(19))/(w+16.3xx10^(-19))` or `w=11xx10^(-19)J=6.875eV` |
|