Concepts Of Physics MCQ Edition [Volume 2]
- Exam
- Concepts Of Physics MCQ Edition [Volume 2]
- Subject
- Physics
- Chapter
- Semiconductors and Semiconductor Devices
Concepts Of Physics MCQ Edition [Volume 2] · Physics · Semiconductors and Semiconductor Devices
Determine the maximum wavelength of electromagnetic radiation...
Determine the maximum wavelength of electromagnetic radiation capable of creating a hole-electron pair in germanium. The band gap of germanium is .
Options
- A
- B
- C
- D
Correct answer
B.
Step-by-step solution
The energy required to create a hole-electron pair is equal to the band gap energy .
The maximum wavelength \lambda_\max of the electromagnetic radiation is given by:
\lambda_\max = \dfrac{hc}{E_g}
Converting the band gap energy into Joules:
Substituting the values of Planck's constant and speed of light :
\lambda_\max = \dfrac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.04 \times 10^{-19}} \text{ m}
\lambda_\max = \dfrac{19.89 \times 10^{-26}}{1.04 \times 10^{-19}} \text{ m}
\lambda_\max \approx 19.1 \times 10^{-7} \text{ m}
\lambda_\max \approx 1.9 \times 10^{-6} \text{ m}
The maximum wavelength \lambda_\max of the electromagnetic radiation is given by:
\lambda_\max = \dfrac{hc}{E_g}
Converting the band gap energy into Joules:
Substituting the values of Planck's constant and speed of light :
\lambda_\max = \dfrac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.04 \times 10^{-19}} \text{ m}
\lambda_\max = \dfrac{19.89 \times 10^{-26}}{1.04 \times 10^{-19}} \text{ m}
\lambda_\max \approx 19.1 \times 10^{-7} \text{ m}
\lambda_\max \approx 1.9 \times 10^{-6} \text{ m}