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Home » A p-n photodiode is fabricated from a semiconductor with a bandgap of Can it detect a wavelength of 6000 nm?
A p-n photodiode is fabricated from a semiconductor with a bandgap of 2.8 \mathrm{eV} . Can it detect a wavelength of 6000 nm?
CBSE | Class 12 | NCERT | Physics | Semiconductor Electronics: Materials, Devices and Simple Circuits
A p-n photodiode is fabricated from a semiconductor with a bandgap of 2.8 \mathrm{eV} . Can it detect a wavelength of 6000 nm?

No, the photodiode is unable to detect wavelengths of 6000 nm due to the following:

The energy bandgap of the given photodiode is given as E_{g}=2.8 \mathrm{eV}

The wavelength has the value \lambda=6000 \mathrm{~nm}=6000 \times 10^{-9} \mathrm{~m}

The energy of the signal is represented by the following relation:

E=h c / \lambda

In the equation, h is the Planck’s constant \left.=6.626 \times 10^{-34}\right\rfloor and c is the speed of light =3 \times 10^{8} \mathrm{~m} / \mathrm{s}.

Putting the values in the equation, we get

E=\left(6.626 \times 10^{-34} \times 3 \times 10^{8}\right) / 6000 \times 10^{-9}=3.313 \times 10^{-20} \mathrm{~J}

But, 1.6 \times 10^{-19} \mathrm{~J}=1 \mathrm{eV}

Therefore, E=3.313 \times 10^{-20} \mathrm{~J}=3.313 \times 10^{-20} / 1.6 \times 10^{-19}=0.207 \mathrm{eV}

A signal with a wavelength of 6000 nm has an energy of 0.207eV, which is less than the energy band gap of a photodiode, which is 2.8eV. As a result, the photodiode is unable to detect the signal.

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