Question

The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than 600 nm is incident on it. The energy band gap (in eV) for the semiconductor is

• A. 1.50
• B. 0.75
• C. 2.06
• D. 1.35
• E. 0.90

Hint:

Use \(E(eV)=1240/\lambda(nm)\) for quick calculation.

Answer 1

The Correct Option is C

Understanding the Concept: For photo-excitation in semiconductors: \[ E_g = \frac{hc}{\lambda} \]

Step 1: Identify threshold wavelength.
\[ \lambda = 600\,nm = 600 \times 10^{-9} m \]

Step 2: Use formula in eV. \[ E_g = \frac{1240}{\lambda (nm)} \]

Step 3: Substitute. \[ E_g = \frac{1240}{600} = 2.066\,eV \]

Step 4: Final Answer. \[ \boxed{2.06\,eV} \]