Ch 42: Molecules and Condensed Matter

Young & Freedman Calc - University Physics 15th Edition

Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook

Young & Freedman Calc 15th Edition

Ch 42: Molecules and Condensed Matter

Problem 26a

Chapter 41, Problem 26a

Suppose a piece of very pure germanium is to be used as a light detector by observing, through the absorption of photons, the increase in conductivity resulting from generation of electron–hole pairs. If each pair requires $0.67$ eV of energy, what is the maximum wavelength that can be detected? In what portion of the spectrum does it lie?

Verified step by step guidance

Step 1: Understand the relationship between energy and wavelength. The energy of a photon is given by the equation:

E = h c / λ

, where

E

is the energy of the photon,

h

is Planck's constant (

6.626 \times 10 ⁻ 34

J·s),

c

is the speed of light (

3.00 \times 10 ⁸

m/s), and

λ

is the wavelength of the photon.

Step 2: Convert the energy required to generate an electron-hole pair from electron volts (eV) to joules (J). Use the conversion factor:

1 eV = 1.602 \times 10 ⁻ 19 J

. Multiply

0.67 eV

by this factor to find the energy in joules.

Step 3: Rearrange the photon energy equation to solve for wavelength:

λ = h c / E

. Substitute the values for

h

c

, and the energy (in joules) into this equation.

Step 4: Perform the calculation to determine the maximum wavelength

λ

. Ensure that the units are consistent (meters for wavelength).

Step 5: Identify the portion of the electromagnetic spectrum corresponding to the calculated wavelength. Compare the wavelength to the ranges of the spectrum (e.g., ultraviolet, visible, infrared) to determine where it lies.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Photon Energy and Wavelength Relationship

The energy of a photon is inversely related to its wavelength, described by the equation E = hc/λ, where E is energy, h is Planck's constant, c is the speed of light, and λ is the wavelength. This relationship indicates that shorter wavelengths correspond to higher energy photons, which is crucial for determining the maximum wavelength that can generate electron-hole pairs in germanium.

Electron-Hole Pair Generation

In semiconductors like germanium, when a photon with sufficient energy is absorbed, it can excite an electron from the valence band to the conduction band, creating an electron-hole pair. This process is fundamental to the operation of light detectors, as the generation of these pairs leads to increased conductivity, allowing the detection of light.

Energy Bands in Semiconductors

Semiconductors have distinct energy bands: the valence band, filled with electrons, and the conduction band, where electrons can move freely. The energy gap between these bands determines the minimum energy required to generate electron-hole pairs. For germanium, this gap is approximately 0.67 eV, which is essential for calculating the maximum detectable wavelength of light.

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A forward-bias voltage of $15.0$ mV produces a positive current of $9.25$ mA through a $p - n$ junction at $300$ K. What does the positive current become if the forward-bias voltage is reduced to $10.0$ mV?

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The maximum wavelength of light that a certain silicon photocell can detect is 1.11 mm. (b) Explain why pure silicon is opaque.

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Textbook Question

At the Fermi temperature $T_{F}$ , $E_{F} = k T_{F}$ (see Exercise $42.22$ ). When $T = T_{F}$ , what is the probability that a state with energy $E equals 2 E sub F$ is occupied?

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Textbook Question

Pure germanium has a band gap of $0.67$ eV. The Fermi energy is in the middle of the gap. For temperatures of $250$ K, $300$ K, and $350$ K, calculate the probability $f (E)$ that a state at the bottom of the conduction band is occupied.

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