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

Explanation

Energy band gap of the given photodiode, `E_(g) = 2.8 eV`
Wavelength, `lamda = 6000 nm = 6000 × 10^(−9) m`
The energy of a signal is given by the relation:
`E=overset((hc)/(lamda)`
Where,
h = Planck’s constant
`= 6.626 xx 10−^(34) "Js"`
c = Speed of light
`= 3 × 10^(8) " m/s"`
`=(6.626xx10^(-34)xx3xx10^(8))/(600xx10^(-9))`
E
`= 3.313 xx 10^(−20) J`
But `1.6 xx 10^(−19) J = 1 eV`
`thereforeE = 3.313 xx 10^(−20) J`
`=(3.313xx10^(-20))/(1.6xx10^(-19))=0.207 " eV"`
The energy of a signal of wavelength 6000 nm is 0.207 eV, which is less than 2.8 eV − the energy band gap of a photodiode. Hence, the photodiode cannot detect the signal.