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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

`172220Å`

B

`172.2Å`

C

`17222Å`

D

`1722Å`

Text Solution

Verified by Experts

The correct Answer is:
C

Energy gap , `E_g=(hc)/lamda`
`lamda=(hc)/E_g=(12400)/(0.72)=17222Å`
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