Question

The band gap between conduction band and valence bond for germanium is 0.66 eV. Find the range for wavelength of radiation that can cause generation of hole-electron pair.

• A. 2.1 x 10-6 m and 2.3 x 10-6 m
• B. 1.7 x 10-6 m and 2.2 x 10-6 m
• C. 1.6 x 10-6 m and 1.2 x 10-6 m
• D. 1.5 x 10-6 m and 2.9 x 10-6 m

Answer 1

Answer: (B)

1.7 x 10-6 m and 2.2 x 10-6 m

Answer 2

The energy required to generate a hole-electron pair is equal to the band gap energy ($E_g$). The energy of a photon is related to its wavelength ($\lambda$) by the equation: $E = \frac{hc}{\lambda}$

For a photon to cause the generation of a hole-electron pair, its energy must be at least equal to the band gap energy: $E_{photon} \ge E_g$ $\frac{hc}{\lambda} \ge E_g$

This implies that the wavelength of the photon must be less than or equal to a maximum value, $\lambda_{max}$: $\lambda \le \frac{hc}{E_g}$

Given: Band gap energy of Germanium, $E_g = 0.66$ eV. Planck's constant, $h = 6.62 \times 10^{-34}$ J.s Speed of light, $c = 3 \times 10^8$ m/s Conversion factor, $1$ eV $= 1.6 \times 10^{-19}$ J

Convert the band gap energy to Joules: $E_g = 0.66 \text{ eV} \times (1.6 \times 10^{-19} \text{ J/eV}) = 1.056 \times 10^{-19}$ J

Calculate the maximum wavelength ($\lambda_{max}$): $\lambda_{max} = \frac{hc}{E_g} = \frac{(6.62 \times 10^{-34} \text{ J.s}) \times (3 \times 10^8 \text{ m/s})}{1.056 \times 10^{-19} \text{ J}}$ $\lambda_{max} \approx 1.88 \times 10^{-6}$ m

Therefore, the wavelength of incident radiation must be less than or equal to approximately $1.88 \times 10^{-6}$ m. The range of wavelengths that can cause this effect is $(0, 1.88 \times 10^{-6}]$ m.

Among the given options, option B, "1.7 x 10-6 m and 2.2 x 10-6 m", is the most appropriate. This range includes wavelengths that can cause pair generation, with the critical wavelength $1.88 \times 10^{-6}$ m falling within this interval. The portion of this range that can cause generation is $[1.7 \times 10^{-6} \text{ m}, 1.88 \times 10^{-6} \text{ m}]$.