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The forbidden energy gap in Ge in 0.72 e...

The forbidden energy gap in Ge in 0.72 eV Given, `hc=12400" eV"-Å` The maximum wavelength of radiation that will generate electron hole pair is

A

`1722Å`

B

`17222Å`

C

`172.2Å`

D

`17220Å`

Text Solution

Verified by Experts

The correct Answer is:
D
`E=(hc)/lamda` , then `lamda=(hc)/E=12400/(0.27)=17220 Å`
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MODERN PUBLICATION-SOLIDS & SEMICONDUCTOR DEVICES-Revision Test
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  3. Choose the reverse biased p-n junction :

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