To detect light of wavelength 500 nm , the photodiode must be fabricated from a semiconductor of minimum bandwidth of

Question

To detect light of wavelength 500 nm , the photodiode must be fabricated from a semiconductor of minimum bandwidth of (A) 1.24 eV (B) 0.62 eV (C) 2.48 eV (D) 3.2 eV

Tardigrade Admin · Accepted Answer

Minimum bandwidth Δ E Δ E=(1240 eV nm/500 nm)=2.48 eV

Q. To detect light of wavelength $500 nm$ , the photodiode must be fabricated from a semiconductor of minimum bandwidth of

$1.24 e V$

$0.62 e V$

$2.48 e V$

$3.2 e V$

Minimum bandwidth $Δ E$
$Δ E = \frac{1240 e Vnm}{500 nm} = 2.48 e V$

Q. To detect light of wavelength 500nm , the photodiode must be fabricated from a semiconductor of minimum bandwidth of

1.24eV

0.62eV

2.48eV

3.2eV