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The width of a depletion region is 400 n...

The width of a depletion region is 400 nm. The intensity of the electric field at the depletion region is `5xx10^(5)`V/m. Then calculate the following quantities: (1) The value of the potential barrier. (2) The minimum energy required by an electron to move from the n-type to the p-type region of the diode.

Text Solution

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`d=400 nm = 400xx10^(-9)m`
`E=5xx10^(5)V//m`
`1eV=1.6xx10^(-19)J`
(1) `E=(V)/(d)`
`therefore V=E.d=5xx10^(5)xx400xx10^(-9)`
`therefore V = 0.2V`
(2) Kinetic energy `(1)/(2)mv^(2)=Ve`
`=0.2xx1.6xx10^(-19)`
`=3.2xx10^(-20)J`
` =(3.2xx10^(-20))/(1.6xx10^(-19))`
Kinetic energy `=0.2eV`
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