Question

To detect light of wavelength $500$ nm, the photodiode must be fabricated from a semiconductor of minimum band gap of

• A. $1.24$ eV
• B. $0.62$ eV
• C. $2.48$ eV
• D. $3.2$ eV
• E. $4.48$ eV

Hint:

For photodetectors: $E_g \leq \frac{1240}{\lambda}$ — choose smallest valid option.

Answer 1

The Correct Option is B

Concept: Photon energy and band gap condition
For a photodiode to detect light: \[ \text{Photon energy} \geq \text{Band gap energy} \] \[ E \geq E_g \] ---

Step 1: Calculate photon energy
Using standard relation: \[ E = \frac{hc}{\lambda} \] Shortcut (in eV): \[ E(\text{eV}) = \frac{1240}{\lambda (\text{nm})} \] ---

Step 2: Substitute wavelength \[ E = \frac{1240}{500} \] \[ E = 2.48 \text{ eV} \] ---

Step 3: Apply detection condition
For detection: \[ E \geq E_g \] So: \[ E_g \leq 2.48 \text{ eV} \] ---

Step 4: Understand what “minimum band gap” means

• Question asks: minimum band gap required
• Any semiconductor with $E_g \leq 2.48$ eV will detect the light
• Among given options, we must choose the smallest possible value ---

Step 5: Compare options

• $0.62$ eV → satisfies $E_g \leq 2.48$ ✔
• $1.24$ eV → also valid but not minimum
• $2.48$ eV → borderline case
• Larger values → cannot detect ---

Step 6: Physical Interpretation

• Lower band gap → easier electron excitation
• Photon energy creates electron-hole pair
• This generates current → detection --- Final Answer: \[ \boxed{0.62 \text{ eV}} \]