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JEE Main 2024 (6 Apr, Shift 2) · Intrinsic & Extrinsic Semiconductors

Intrinsic & Extrinsic Semiconductors question from JEE Main 2024 (6 Apr, Shift 2)

Q23JEE Main 2024 · 6 APR · SHIFT 2Medium

Intrinsic & Extrinsic Semiconductors

The acceptor level of a p-type semiconductor is $6\text{ eV}$ . The maximum wavelength of light which can create a hole would be: Given $hc = 1242\text{ eV nm}$ .

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Answer: (d)

To create a hole, a photon must lift a valence electron to the acceptor level: $E = \dfrac{hc}{\lambda} \ge 6$ eV.
$\lambda_{max} = \dfrac{hc}{E} = \dfrac{1242}{6}$ nm
$\lambda_{max} = 207$ nm
So the answer is (d) 207 nm.

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