A `p-n` photodiode is fabricated from a semiconductor with a band gap of `2.5 eV`. It can detect a signal of wavelength

Options

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

Answer

D

Explanation

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