36. A p-n photodiode is made of a material with a band gap of 2.0eV. The minimum frequency of the radiation that can be absorbed by the material is nearly ?
\( (a)\ 10 \times 10^{14}\ \text{Hz} \)
\( (b)\ 5 \times 10^{14}\ \text{Hz} \)
\( (c)\ 1 \times 10^{14}\ \text{Hz} \)
\( (d)\ 20 \times 10^{14}\ \text{Hz} \)
Answer : \( (b)\ 5 \times 10^{14}\ \text{Hz} \)
\( E_{\text{photon}} = h\nu = \frac{hc}{\lambda} (\text{eV}) \)
\( E = 2.0 \times (1.6 \times 10^{-19}) = 3.2 \times 10^{-19}\ \text{J} \)
\( E = h\nu \)
\( \nu = \frac{E}{h} \)
\( \nu = \frac{3.2 \times 10^{-19}}{6.626 \times 10^{-34}} \)
\( \nu \approx 4.83 \times 10^{14}\ \text{Hz} \)
\( \nu \approx 5 \times 10^{14}\ \text{Hz} \)
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