The potentiaI barrier of a p-n junction diode is 0.4 V . If the thickness of the depIetion region be , what wiII be `4.0 xx 10^(-7)`the eIectric fieId intensity in this region? An eIectron from the n-region moves towards the p-n junction with veIocity `6xx10^(5) "m".s^(-1)`. What will be the veIocity of the eIec- tron with which it enters the p-region?

Explanation

Electric field intensity `E =(V)/(d)`
Here V= value of potential barrier =0.4 V
and d= thickness of depletion region `=4xx10^(-7)`m.
`:. " "E=(0.4)/(4xx10^(-7))=10^(6 V. "m"^(-1)`
Let an electron enters to the depletion region from n-region with velocity `v_(1)` and come out from the depletion region with velocity `v_(2)`. Due to this increase in potential energy is eV.
`:.` Accordin to the principle of conservation of energy,
`(1)/(2)"mv"_(1)^(2)=eV+(1)/(2)"mv"_(2)^(2)`
or `(1)/(2) (9.1xx10^(-31)) xx(6xx10^(5))^(2)`
`=1.6 xx10^(-19)xx0.4+(1)/(2)xx(9.1xx10^(-31))."v"_(2)^(2)`
or `1.64xx10^(-19)=0.64xx10^(-19)+4.55xx10^(-31)."v"_(2)^(2)`
or `"v"_(2)^(2)=(1xx10^(-19))/(4.55xx10^(-31))=22xx10^(10)`
or `"v"_(2)=4.7xx10^(5)"m".s^(-1)`