Question

What is the minimum wavelength of radiation required to detect a p-n junction diode made of a semiconductor having a band gap of $3\text{ eV}$? (Take Planck's constant $h = 6 \times 10^{-34}\text{ J s}$, speed of light $c = 3 \times 10^8\text{ m s}^{-1}$)}

• A. $3300\text{ }^\circ\text{A}$
• B. $4800\text{ }^\circ\text{A}$
• C. $3750\text{ }^\circ\text{A}$
• D. $7500\text{ }^\circ\text{A}$

Hint:

To bypass intermediate unit conversions, use the shortcut relation $\lambda \text{ (in }^\circ\text{A)} \approx \frac{12400}{E_g \text{ (in eV)}}$. Substituting $3.3\text{ eV}$ directly yields $\frac{12400}{3.3} \approx 3757.5\text{ }^\circ\text{A}$, pointing cleanly to $3750\text{ }^\circ\text{A}$.

Answer 1

The Correct Option is C

Understanding the Concept: For a semiconductor device to absorb a photon and excite an electron across its energy band gap ($E_g$), the energy carried by the incident photon must be at least equal to or greater than the band gap value ($E_{\text{photon}} \ge E_g$). The relationship between threshold photon energy and its wavelength ($\lambda$) is expressed by: \[ E_g = \frac{hc}{\lambda} \implies \lambda = \frac{hc}{E_g} \]
Step 1: Convert the band gap energy value into standard SI units (Joules).
Given $E_g = 3.3\text{ eV}$: \[ E_g = 3.3 \times 1.6 \times 10^{-19}\text{ J} = 5.28 \times 10^{-19}\text{ J} \]
Step 2: Calculate the wavelength value.
Using the isolated wave equation parameters: \[ \lambda = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{5.28 \times 10^{-19}} \] \[ \lambda = \frac{1.98 \times 10^{-25}}{5.28 \times 10^{-19}} = 0.375 \times 10^{-6}\text{ m} \] Converting meters to Angstrom units ($1\text{ }^\circ\text{A} = 10^{-10}\text{ m}$): \[ \lambda = 3750 \times 10^{-10}\text{ m} = 3750\text{ }^\circ\text{A} \]