Saved Bookmarks
| 1. |
The resostivity of an impurity-free semiconductor at room tempterature is rho=50 Omega cm. It becomes equal to rho_(1)= 40 Omega cm. When the semiconductor is illuminated with light, and t= 8ms after swithching off the light source the resistivity becomes equal to rho_(2)= 45 Omega cm. Find the mean lifetime of conduction electrons and holes. |
|
Answer» Solution :We write the CONDUCTIVITY of the simple as `sigma= sigma_(i)+sigma_(gamma)` where `sigma_(i)=` intrinsic conductivity and `sigma_(gamma)` is the photo conductivity. At `t=0`, assuming saturation we have `(1)/(rho_(1))=(1)/(rho)+ sigma_(gamma_(0)) or sigma_(gamma_(0))=(1)/(rho_(1))-(1)/(rho)` Time `t` after light source is switched off we have beacuse of recombination of electron and holes in the sample `sigma= sigma_(i)+sigma_(gamma_(0))e^(-t//T)` where `T=` mean lifetime of electrons and holes. Thus `Thus (1)/(rho_(2))=(1)/(rho)+((1)/(rho_(1))-(1)/(rho))e^(-t//T)` or `(1)/(rho_(2))-(1)/(rho)=((1)/(rho_(1))-(1)/(rho))e^(-t//T)` or `e^(t//T)=((1)/rho_(1)-(1)/(rho))/((1)/(rho_(2))-(1)/(rho))=(rho_(2)(rho-rho_(1)))/(rho_(1)(rho-rho_(2)))` Hence `T= t//In {(rho_(2)(rho-rho_(1)))/(rho_(1)(rho-rho_(2)))}` Substitution gives `T= 9.87 ms~ 0.01 sec` |
|