Question

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
\([h=6.6\times 10^{-34}\text{Js},C=3\times 10^{8}\text{ m/s}]\)

• A. \(6000\) Å
• B. \(4000\) Å
• C. \(6000\) nm
• D. \(4000\) nm

Hint:

A photodiode detects light only if the photon energy is at least the band gap, so find the cut-off wavelength using hc over the band gap.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
A photodiode responds when photon energy \(\dfrac{hc}{\lambda}\ge E_g\), so it detects wavelengths shorter than a cut-off.

Step 2: Calculation
\(\lambda_{max}=\dfrac{hc}{E_g}=\dfrac{6.6\times10^{-34}\times3\times10^{8}}{4\times10^{-19}}\)
\(=4.95\times10^{-7}\text{ m}=4950\text{ Å}\).

Step 3: Compare the options
Detectable wavelengths must be below 4950 Å.
6000 Å, 4000 nm and 6000 nm are all longer than this.

Step 4: Conclusion
Only 4000 Å is shorter than the cut-off.

Final Answer:
The photodiode can detect light of wavelength 4000 Å. \[ \boxed{\text{(B) }\text{4000 Å}} \]