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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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Knowledge Check

  • The forbidden energy gap in Ge is 0.72 eV. Given hc=12400 eV - Å . The maximum wavelength of radiation that will generate an electron hole pair is :

    A
    `1722Å`
    B
    `17222Å`
    C
    `172.2Å`
    D
    `17220Å`

  • Find the de-Broglie wavelength of an electron with kinetic energy of 120 eV.

    A
    102 pm
    B
    124 pm
    C
    95 pm
    D
    112 pm

  • Find the de-Broglie wavelength of an electron with kinetic energy of 120 eV.

    A
    102
    B
    124 pm
    C
    95 pm
    D
    112 pm

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