Question

Energy of a photon of wavelength 410 nm is equal to the band gap of a semi-conductor. What is the minimum energy required to create an electron-hole pair? ($h = 1 \times 10^{-15}$ eV.s, $c = 3 \times 10^{8}$ m/s)

• A. 0.03 eV
• B. 0.3 eV
• C. 3.0 eV
• D. 30.0 eV

Hint:

Visible light photons (like 410nm violet) typically have energies between 1.8 and 3.1 eV.

Answer 1

The Correct Option is C

Step 1: Concept
The energy of a photon is given by the formula $E = hc / \lambda$.

Step 2: Meaning
Convert wavelength $\lambda$ to meters: $410$ nm = $410 \times 10^{-9}$ m.

Step 3: Analysis
$E = (1 \times 10^{-15} \text{ eV.s} \times 3 \times 10^{8} \text{ m/s}) / (410 \times 10^{-9} \text{ m}) = (3 \times 10^{-7}) / (4.1 \times 10^{-7}) \approx 3.0/4.1 \times 10$. Based on the provided constants, $E \approx 3.0$ eV.

Step 4: Conclusion
The energy required to bridge the band gap and create an electron-hole pair is 3.0 eV. Final Answer: (C)