A p-n junction photodiode is fabricated from a semiconductor with a band gap of 2.5 eV. It can detect a signal of wavelength [Planck's constant = 6.6 × 10⁻³⁴ Js, c = 3 × 10⁸ m/s, e = 1.6 × 10⁻¹⁹ C ] (a) 6000 nm (b) 6000 (c) 5000 (d) 4000 nm

Answer: (c). A photodiode detects light only if photon energy ≥ Eg, so λmax = hc/Eg. λmax = 6.6×10⁻³⁴×3×10⁸2.5×1.6×10⁻¹⁹ = 4.95×10⁻⁷ m ≈ 5000 Å 6000 Å, 6000 nm and 4000 nm are all longer than this cutoff, so they cannot be detected. So the answer is (c) 5000 Å.

A p-n junction photodiode is fabricated from a semiconductor with a… | MHT CET 2021 (24 Sep, Shift 1)

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Special Purpose Diodes

A p-n junction photodiode is fabricated from a semiconductor with a band gap of $2.5\text{ eV}$ . It can detect a signal of wavelength [Planck's constant $= 6.6 \times 10^{-34}\text{ Js}, \text{c} = 3 \times 10^{8}\text{ m/s}, \text{e} = 1.6 \times 10^{-19}\text{ C}$ ]

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Answer: (c)

A photodiode detects light only if photon energy $\geq E_g$ , so $\lambda_{max} = \dfrac{hc}{E_g}$ .
$\lambda_{max} = \dfrac{6.6\times10^{-34}\times3\times10^{8}}{2.5\times1.6\times10^{-19}} = 4.95\times10^{-7}\text{ m} \approx 5000\text{ Å}$
6000 Å, 6000 nm and 4000 nm are all longer than this cutoff, so they cannot be detected.
So the answer is (c) 5000 Å.

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