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

Explanation

`E_(E) = 2.5 eV = 2.5 xx 1.6 xx 10^(-19) J`
`(hc)/( lambda_(max)) = 4 xx 10^(19) J`
`=4. 9695 xx 10^(-7) m = 4969 .5 xx 10^(-10) m`
Hence diode can detect signal of wavelength less than calculated `lambda_(max)`
most appropriate signal is 400 A