At T = 300 K, the bandgap and the intrinsic carrier concentration
At T = 300 K, the bandgap and the intrinsic carrier concentration of GaAs are 1.42 eV and 106 cm-3, respectively. In order to generate electron-hole pairs in GaAs, which one of the wavelength (\({{\rm{\lambda }}_c}\)) ranges of incident radiation, is most suitable?
(Given that: Plank’s constant is 6.62 × 10-34 J-s, the velocity of light is 3 × 1010 cm/s and charge of an electron is 1.6 × 10-19 C)
A. 0.42 μm < <span class="math-tex">\({{\rm{\lambda }}_c}\)</span>< 0.87 μm
B. 0.87 μm < <span class="math-tex">\({{\rm{\lambda }}_c}\)</span>< 1.42 μm
C. 1.42 μm < <span class="math-tex">\({{\rm{\lambda }}_c}\)</span>< 1.62 μm
D. 1.62 μm < <span class="math-tex">\({{\rm{\lambda }}_c}\)</span>< 6.62 μm
Please scroll down to see the correct answer and solution guide.
Right Answer is: A
SOLUTION
Concept:
The frequency and wavelength are related by the relation:
\(\nu =\frac{c}{λ}\)
Less the wavelength more will be the frequency and vice-versa.
Also, the energy of a photon with wavelength λ is given by:
\(E= \frac{{hc}}{\lambda}\)
To generate electron-hole pairs, the energy of the incident photon must be greater than the Band-gap energy.
Application:
The wavelength of the incident radiation of energy 1.42 eV will be:
\(λ = \frac{{hc}}{E}\)
\( \lambda= \frac{{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{1.42 \times 1.6 \times {{10}^{ - 19}}}}\)
\(\lambda = 0.87 \times {10^{ - 6}}\;m\)
\(\lambda = \;0.87\;\mu m\)
So, the radiation of wavelength less than 0.87 μm can generate electron-hole pair in Ga As, as the frequency will be more.
∴ Option (1) best satisfies the required condition.