Question

A photo director area light of wavelength 1400 nm Band gap of the Semiconductor used in the photo detector is _______ (h = 6.63 × 10^–34 JS; C = 3 × 10⁸ m/s) 

(A) 0.7 eV 

(B) 1 eV 

(C) 2 eV 

(D) 2.5 eV

Answer 1

The correct option is (B) 1 eV.

Explanation:

λ = 1400 nm = 1400 × 10^–9m 

Band gap = hf = (hc / λ) 

∴ E_g = [(6.625 × 10^–34 × 3 × 10⁸) / (1400 × 10^–9)] 

= 0.0141 × 10^–17 

E_g = 1.41 × 10^–19 

∴ E_g in eV = [(1.41 × 10^–19) / (1.6 × 10^–19)]eV 

= 0.89 

≈ 1 eV