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### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# When a photon enters the depletion zone of a p-n junction, the photon can scatter from the valence electrons there, transferring part of its energy to each electron, which then jumps to the conduction band. Thus, the photon creates electron–hole pairs. For this reason, the junctions are often used as light detectors, especially in the x-ray and gamma-ray regions of the electromagnetic spectrum. Suppose a single 662keV gamma-ray photon transfers its energy to electrons in multiple scattering events inside a semiconductor with an energy gap of 1.1eV, until all the energy is transferred. Assuming that each electron jumps the gap from the top of the valence band to the bottom of the conduction band, find the number of electron – hole pairs created by the process.

The number of electron-hole pairs created by the process is $6×{10}^{5}$.

See the step by step solution

## Step 1: The given data

1. Energy transferred by the gamma-ray photon to the conduction electron $E=662\mathrm{keV}$
2. Energy gap of the semiconductor, ${E}_{g}=1.1\mathrm{eV}$

## Step 2: Understanding the concept of band gap energy

The valence band and the conduction band are separated by a certain amount of energy which is called the band gap energy. Thus, this energy is required by each electron on the top of the valence band to jump to the bottom of the conduction band to make the transition in the valence band. Since the energy received by each electron is exactly the value of band gap energy, thus, the number of electrons that can be excited across the gap by a single photon of a given energy is given by the ratio of the photon energy (E) and the gap energy (Eg).

Formula:

The number of conduction electrons to jump from the top of the valence band to the bottom of the conduction band, $N=E/{E}_{g}$ (i)

## Step 3: Calculation of the number of electron-hole pairs

From the given concept, the number of conduction electrons can be calculated using the given data in equation (i) as -

$N=662\mathrm{keV}/1.1\mathrm{eV}\phantom{\rule{0ex}{0ex}}=6×{10}^{5}$

Since each electron that jumps the gap leaves a hole behind, the above value of conduction electrons also gives us the number of electron-hole pairs.

Hence, the value of the electron-hole pairs is $6×{10}^{5}$ .

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