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A p-n photodiode is fabricated from a se...

A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 6000 nm?

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The radiation energy of the wavelength of 6000 nm.
`E=(hc)/(lambda)=(6.625xx10^(-34)xx3xx10^(8))/(6000xx10^(-9)xx1.6xx10^(-19))eV`
`=0.00207xx10^(2)eV`
`=0.207eV`
but `E_(g)=2.8eV` is given
`therefore` So, `E gt E_(g)` should be done to test the wavelength of radiation but since there is `E lt E_(g)` the wavelength cannot be tested.
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