Q200 / 200 · MHT CET - Medical 2026 (22 April Shift 1) · Physics
A p-n junction photodiode is fabricated from a semiconductor with a band gap of $4 \times 10^{-19}$ J. It can detect a signal of wavelength
$[h = 6.6 \times 10^{-34} \text{Js} , C = 3 \times 10^8 \text{ m/s}]$
$[h = 6.6 \times 10^{-34} \text{Js} , C = 3 \times 10^8 \text{ m/s}]$
- A$6000$ Å
- Correct answerB$4000$ Å
- C$6000$ nm
- D$4000$ nm
Solution
For a photodiode to detect a signal, the energy of the incident photon must be greater than or equal to the band gap energy of the semiconductor.
$E \ge E_g$
$\dfrac{hc}{\lambda} \ge E_g$
$\lambda \le \dfrac{hc}{E_g}$
Substituting the given values:
$\lambda \le \dfrac{6.6 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^{-19}}$
$\lambda \le \dfrac{19.8 \times 10^{-26}}{4 \times 10^{-19}}$
$\lambda \le 4.95 \times 10^{-7} \text{ m}$
$\lambda \le 4950 \text{ \AA}$
Among the given options, only $4000 \text{ \AA}$ is less than $4950 \text{ \AA}$.
Answer: $4000$ Å
$E \ge E_g$
$\dfrac{hc}{\lambda} \ge E_g$
$\lambda \le \dfrac{hc}{E_g}$
Substituting the given values:
$\lambda \le \dfrac{6.6 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^{-19}}$
$\lambda \le \dfrac{19.8 \times 10^{-26}}{4 \times 10^{-19}}$
$\lambda \le 4.95 \times 10^{-7} \text{ m}$
$\lambda \le 4950 \text{ \AA}$
Among the given options, only $4000 \text{ \AA}$ is less than $4950 \text{ \AA}$.
Answer: $4000$ Å
Hence, the correct answer is option: 'B'.