Question

The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than \(1.24\,\mu m\) is incident on it. The band gap (in eV) for the semiconductor is

• A. \(1\,eV\)
• B. \(1.1\,eV\)
• C. \(2.48\,eV\)
• D. \(0.7\,eV\)

Hint:

Use shortcut: \(E(\text{eV}) = \frac{1.24}{\lambda(\mu m)}\) for quick band gap calculations.

Answer 1

The Correct Option is A

Step 1: Concept of band gap and photon energy.
Electrical conductivity increases when photon energy is equal to or greater than band gap energy.
\[ E = \frac{hc}{\lambda} \]

Step 2: Use standard relation.
In electron volts:
\[ E(\text{eV}) = \frac{1.24}{\lambda(\mu m)} \]

Step 3: Substitute given wavelength.
\[ E = \frac{1.24}{1.24} \]
\[ E = 1\,eV \]

Step 4: Interpretation.
This is the minimum photon energy required to excite electrons across the band gap.

Step 5: Final conclusion.
\[ \boxed{1\,eV} \] Hence, correct answer is option (A).