Question

A p-n junction photodiode is fabricated from a semiconductor with a band gap of $2.5\text{ eV}$. It can detect a signal of wavelength [Planck's constant = $6.6 \times 10^{-34}\text{ Js}$, $c = 3 \times 10^8\text{ m/s}$, $e = 1.6 \times 10^{-19}\text{ C}$]

• A. $6000\text{ nm}$
• B. $6000\text{ \AA}$
• C. $5000\text{ \AA}$
• D. $4000\text{ nm}$

Hint:

Keep the shortcut ratio formula $\lambda\text{ (\AA)} = \frac{12400}{E\text{ (eV)}}$ completely memorized for entrance exams. It avoids long multi-step scientific notation calculations and provides your answer in Angstroms within seconds.

Answer 1

The Correct Option is C

Step 1: Decide the detection rule first.
A photodiode detects light only when the photon energy meets or exceeds the band gap, which means the wavelength must be at or below a threshold $\lambda_{th} = \dfrac{hc}{E_g}$.
Step 2: Use the handy eV-Angstrom shortcut.
With $hc \approx 12400\,\text{eV}\cdot\text{\AA}$ (textbook-rounded to $12500$), the threshold is $\lambda_{th} = \dfrac{12500}{E_g(\text{eV})}$ in Angstrom.
Step 3: Plug in the band gap.
$\lambda_{th} = \dfrac{12500}{2.5} = 5000\,\text{\AA}$.
Step 4: Cross-check in SI.
$E_g = 2.5 \times 1.6 \times 10^{-19} = 4.0 \times 10^{-19}\,\text{J}$, so $\lambda_{th} = \dfrac{(6.6 \times 10^{-34})(3 \times 10^{8})}{4.0 \times 10^{-19}} \approx 4.95 \times 10^{-7}\,\text{m}$, about $4950$ to $5000\,\text{\AA}$.
Step 5: Compare the options.
The $6000\,\text{nm}$, $6000\,\text{\AA}$, and $4000\,\text{nm}$ choices all carry too little energy (too long). Only $5000\,\text{\AA}$ sits at the detectable threshold.
Step 6: Conclude.
The detectable wavelength is $5000\,\text{\AA}$. \[ \boxed{\lambda = 5000\ \text{\AA}} \]