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The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than 2480 nm is incident on it. The band gap in  ( eV )   for the semiconductor is

  • Option 1)

    0.5 eV

  • Option 2)

    0.7 eV

  • Option 3)

    1.1 eV

  • Option 4)

    2.5 eV

 

Answers (1)

best_answer

As we have learnt,

 

Photo diode -

A special type of photo detector.

- wherein

By measuring the change in the conductance of the semicondutors one can measure the intensity of the optical signal

 

 Band gap = \frac{hc}{\lambda}

                  \\= \frac{12400}{\lambda(\overset {o}A)} = \frac{1240}{\lambda(nm)}eV \\=\frac{1240}{2480}eV = 0.5eV

 


Option 1)

0.5 eV

Option 2)

0.7 eV

Option 3)

1.1 eV

Option 4)

2.5 eV

Posted by

Avinash

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