In zinc oxide (ZnO), the band gap between the valence and conduction bands is 3.2 eV . Assume that an electron in the conduction band recombines with a hole in the valence band, releasing the excess energy as

Correct answer: A. 388 nm The energy of the emitted photon is equal to the band gap energy of the material. E = (hc)/(λ) λ = (hc)/(E) Substituting hc 1242 eV nm and E = 3.2 eV : λ = (1242)/(3.2) nm λ = 388.125 nm 388 nm

Concepts Of Physics MCQ Edition [Volume 2]

Exam: Concepts Of Physics MCQ Edition [Volume 2]
Subject: Physics
Chapter: Semiconductors and Semiconductor Devices

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Semiconductors and Semiconductor Devices

In zinc oxide (ZnO), the band gap between the valence and...

In zinc oxide (ZnO), the band gap between the valence and conduction bands is

3.2 \text{ eV}

. Assume that an electron in the conduction band recombines with a hole in the valence band, releasing the excess energy as electromagnetic radiation. Determine the maximum wavelength of the radiation that can be emitted during this process.

Options

A $388 \text{ nm}$
B $582 \text{ nm}$
C $194 \text{ nm}$
D $776 \text{ nm}$

Correct answer

A. $388 \text{ nm}$

Step-by-step solution

The energy of the emitted photon is equal to the band gap energy of the material.

E = \dfrac{hc}{\lambda}

\lambda = \dfrac{hc}{E}

Substituting

hc \approx 1242 \text{ eV nm}

and

E = 3.2 \text{ eV}

\lambda = \dfrac{1242}{3.2} \text{ nm}

\lambda = 388.125 \text{ nm} \approx 388 \text{ nm}

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