InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
When a surface is irradiated with light of wavelength `4950Å`, a photocurrent appears which vanishes if a retarding potential appears which vanishes if a retarding potential greater than 1.2 volt is applied across the phototube. When a different source of light is used, it is found that the critical retarding potential is changed to 2.1 volt. Find the work function of the emitting surface and the wavelength of second source. If the photoelectrons (after emission from the surface) are subjected to a magnetic field of 10 tesla, what changes will be observed in the above two retarding potentials. Use `h=6.6xx10^(-34)Js , e=1.6xx10^(-19)C`. |
|
Answer» Case (i) `eV_(1)=(hc)/(lambda_(1))-phi_(0) or phi_(0)=(hc)/(lambda_(1))-eV_(1)` `:. phi_(0)=((6.6xx10^(-34))xx(3xx10^(8)))/(4950xx10^(-10))-(1.6xx10^(-19))xx1.2` `=2.08xx10^(-19)J=(2.08xx10^(-19))/(1.6xx10^(-19))eV=1.3eV` Case (ii) `eV_(2)=(hc)/(lambda_(2))-phi_(0) or (hc)/(lambda_(1))-eV_(2)+phi_(0)` `=(1.6xx10^(-19))xx1.2+2.08xx10^(-19)` `=5.44xx10^(-19)J` `lambda_(2)=(hc)/(5.44xx10^(-19)` `=((6.6xx10^(-34))xx(3xx10^(8)))/(5.44xx10^(-19))=3639.7Å` When photoelectrons are subjected to a perpendicular magnetic field, there will be no change in the retarding potential as the speed of the photoelectrons remains unchanged but the photoelectron simply describes a circular path in the field. |
|