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A p-n photodiode is made of a material with a band gap of 2.0 eV. The minimum frequency of the radiation that can be absorbed by the material is nearly :
A
$1 \times 10^{14}$ Hz
B
$20 \times 10^{14}$ Hz
C
$10 \times 10^{14}$ Hz
D
$5 \times 10^{14}$ Hz
Detailed Solution
A photon is absorbed only if its energy is at least equal to the band gap: $h\nu \geq E_g$.
The minimum frequency corresponds to $E = h\nu = 2$ eV.
$2 \times 1.6 \times 10^{-19} = 6.6 \times 10^{-34} \times \nu$
$\nu = \dfrac{3.2 \times 10^{-19}}{6.6 \times 10^{-34}}$
$\nu = 4.85 \times 10^{14}$ Hz $\approx 5 \times 10^{14}$ Hz
So the minimum frequency that can be absorbed is nearly $5 \times 10^{14}$ Hz.
The minimum frequency corresponds to $E = h\nu = 2$ eV.
$2 \times 1.6 \times 10^{-19} = 6.6 \times 10^{-34} \times \nu$
$\nu = \dfrac{3.2 \times 10^{-19}}{6.6 \times 10^{-34}}$
$\nu = 4.85 \times 10^{14}$ Hz $\approx 5 \times 10^{14}$ Hz
So the minimum frequency that can be absorbed is nearly $5 \times 10^{14}$ Hz.