Question

The electrical conductivity of a semiconductor increases when a radiation of wavelength shorter than 1000 nm is incident on it. The band gap of the semiconductor is:
1. 2.4 eV
2. 1.2 eV
3. 3.4 eV
4. 4.0 eV

Answer 1

Correct Answer - Option 2 : 1.2 eV

Concept:

The freq. and wavelength are related by the relation:

\(f = \frac{C}{λ }\)

Also, the energy of a photon with wavelength (λ) is given by:

\(E = \frac{{hc}}{λ }J\)

Where,

h : Planck’s constant = 6.626 × 10-34 Js

c : 3 × 108 m/sec

\(E = \frac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{λ }joule\)

\(\because 1\;eV = 1.602 \times {10^{ - 19}}Joule\)

\(E = \frac{{19.878\; \times\; {{10}^{ - 26}}}}{{1.602\; \times \;{{10}^{ - 19}}λ }}ev\;\;\;\;\;\;\;\;\;\)

\(E \approx \frac{{12400}}{{λ \left( Å \right)}}eV\)      ------(1)

Calculation:

Given:

λ = 1000 nm = 1μm

From equation (1)

\(E= \frac{{12400}}{1\mu m}\)

E = 1.24 eV