Physics Electronics PMDC Conceptual Practice
PMDC Verified Question 295 of 494
What is the maximum (threshold) optical wavelength \( \lambda_{\text{max}} \) that can generate electron-hole pairs in pure Germanium with an energy band gap of \( E_g = 0.70\text{ eV} \)? (Take \( h c \approx 1240\text{ eV}\cdot\text{nm} \))
A
1771 nm (Infrared)
B
1127 nm
C
550 nm
D
3500 nm
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option A: 1771 nm (Infrared)
Concept:

The threshold wavelength for band-to-band carrier generation is given by \( \lambda_{\text{max}} = \frac{h c}{E_g} \).

Formula:

$$\lambda_{\text{max}} = \frac{h c}{E_g} = \frac{1240\text{ eV}\cdot\text{nm}}{0.70\text{ eV}} \approx 1771.4\text{ nm} \approx 1771\text{ nm}$$

Solution:

  • Given: \( E_g = 0.70\text{ eV} \) and \( h c \approx 1240\text{ eV}\cdot\text{nm} \).


  • Cutoff wavelength: \( \lambda_{\text{max}} = \frac{1240}{0.70} \approx 1771\text{ nm} \) (short-wave infrared).


  • Germanium absorbs photons across the visible spectrum and into the infrared up to \( 1771\text{ nm} \).


Why other options are incorrect:

  • Option B: 1127 nm is the threshold cutoff wavelength for Silicon.
  • Option C: 550 nm is in the visible green spectrum, corresponding to \( E_g = 2.25\text{ eV} \).
  • Option D: 3500 nm corresponds to \( E_g = 0.35\text{ eV} \).

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