The width of depletion region in p-n jun...

The width of depletion region in p-n junction diode is `500 nm` and an intrinsic electric field of `6xx 10^(5) Vm^(-1)` is also found to exist in it. What is the kinetic energy which a conduction electron must have in order to diffuse from the n-side to p-side?

0.3eV

0.30eV

0.45eV

0.60eV

Text Solution

Verified by Experts

The correct Answer is:

Here, `E=6xx10^(5)Vm^(-1)`
`d=500xx10^(-9)m`
Here, height of potential barrier,
`V_(B)=Ed=6xx10^(5)xx500xx10^(-9)=0.30V`
Max. K.E. of an electron to cross the potential barrier `eV_(B)=exx(0.30V)=0.30eV`

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The width of the depletion region in a p-n junction diode is 400 nm and an intense electric field of 8 xx 10^5 V/m exists in it. What is the kinetic energy which a conduction electron must have in order to diffuse from then region to p region?

0.16eV

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0. 8 eV

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The width of depletion region in PN–junction diode is 500 nm and an internal electric field is 6 x 10^5 V/m . What is the kinetic energy which a conduction electron must have in order to diffuse from the N side to P side :-

`0.03 eV `

`0.3 eV `

`0.45 eV `

`0.6 eV `

In a p-n junction the depletion region is 400 nm wide and electric field of 5 xx 10^(5) Vm^(-1) exists in it. The minimum energy of a conduction electron, which can diffuse from n-side to the p-side is.

`4 eV`

`5 eV`

`0.4 eV`

`0.2 eV`