1.

A p-n photodiodeis fabricated from a semiconductor with a band gap of 2.5 eV. It can detecta signal of wavelength ……..

Answer»

4000 NM
6000 nm
4000 Å
6000 Å

Solution :4000 Å
`lambda_("max")=(hc)/(E )=(6.6xx10^(-34)xx3xx10^(8))/(2.5xx1.6xx10^(-19))=4950`Å
The wavelength obtained by the PHOTODIODE is less than `lambda_("max")` so it can detected to a SIGNAL having wavelength 4000 Å.


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