A p-n Photodiode is fabricated from a semiconductor with a band gap of 2.5 eV. The signal wavelength is

Options

A. 6000 Å
B. 4000 nm
C. 6000 nm
D. 4960 Å

Answer

D

Explanation

Only signal having wavelength less than threshold wavelength will be detercted.
Energy `E=hv=h(c)/(lamda)implieslamda=(hc)/(E)`
Substituting the value of h, c and E in the above equation
`lamda=(6.6xx10^(-34)xx3xx10^(8))/(2.5xx1.6xx10^(-19))=5000Å`
As `4000Ålt5000Å`
Signal of wavelength `4000Å` can be detected by the photodiode