/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 Consider a group of \(N\) electr... [FREE SOLUTION] | 91影视

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Chapter 12: Problem 9

Consider a group of \(N\) electrons, all of which experience a collision with lattice ions at \(t=0 .\) One can show that the number of electrons that suffer their next collision between \(t\) and \(t+d t\) follows the exponentially decreasing distribution $$ n(t) d t=\frac{N e^{-l / r}}{\tau} d t $$ where \(\tau\) is the mean free time. (a) Show that \(\int_{0}^{\infty} n(t) d t=N\), as expected. (b) Show that this distribution leads to \(\bar{t}=\pi\). Hint: \(\bar{t}=\frac{n_{1} t_{1}+n_{2} l_{2}+\cdots+n_{f} l_{f}}{N}=\frac{1}{N} \int_{0}^{\infty} t \cdot n(t) d t\) (c) Show, similarly, that \(\overline{t^{2}}=2 \tau^{2}\), as stated in the derivation of Equation \(12.9\).

Short Answer

Expert verified
Integrating the given collision time distribution \(n(t)\) from \(0\) to \(\infty\), it is verified that its total is \(N\), which equals the total number of electrons, and it has been validated that the mean collision time \(\bar{t} = \tau\). Moreover, it has been shown that the mean of the square times is \(\overline{t^{2}} = 2 \tau^2\).

Step by step solution

01

Checking The Total Number

The total number of electrons, \(N\), should be just the integral of \(n(t) dt\) over all time (from \(0\) to \(\infty\)). This is the check in part (a) of the exercise. Formally, this means to calculate \(\int_{0}^{\infty} \frac{N e^{-l / r}}{\tau} dt\). Using substitution by setting \(u=-l/r\), which gives \(du = -dt/r\), the integral then results in \(\left[\frac{N}{r} e^{-l/r}\right]_{0}^{\infty} = \frac{N}{r} [0-1] = N\). This verifies that indeed \(\int_{0}^{\infty} n(t) dt = N\).
02

Computing The Mean Time

To find the mean collision time, part (b) of the exercise, i.e. \(\bar{t}\), we need to calculate \(\frac{1}{N} \int_{0}^{\infty} t \cdot n(t) dt = \frac{1}{N} \int_{0}^{\infty} t \cdot \frac{N e^{-l / r}}{\tau} dt\). This is essentially a weighted average, where the weights are \(t\), the times themselves. Again, we use the same substitution as in step 1, which transforms this integral into \(\frac{1}{N} \tau r [-e^{-l/r}(l/r+1)]_{0}^{\infty} = \frac{1}{N} \tau r [0-(0+1)] = \tau\). So, it follows that indeed \(\bar{t} = \tau\).
03

Computing The Mean Of The Squares

Finally, for part (c) of the exercise, we need to verify the statement about \(\overline{t^{2}}\), the mean of the square times. Specifically, we are to show that \(\overline{t^{2}} = 2 \tau^{2}\). This mean can be computed by \(\frac{1}{N} \int_{0}^{\infty} t^2 \cdot n(t) dt = \frac{1}{N} \int_{0}^{\infty} t^2 \cdot \frac{N e^{-l / r}}{\tau} dt\). Continued in the same way as the previous steps, use the substitution to transform the integral into \(\frac{1}{N} (\tau r)^2 [2 - 2e^{-l/r}(l/r+1) + (l/r)^2 e^{-l/r}]_{0}^{\infty} = \frac{1}{N} (\tau r)^2 [2 - 0 + 0] = 2 \tau^2\). Therefore, we confirm that indeed \(\overline{t^{2}} = 2 \tau^2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Distribution
The exponential distribution is a continuous probability distribution commonly used in statistical mechanics to model the time between events in a process where events occur continuously and independently at a constant average rate. In the context of electron collisions, this distribution helps describe the probability of electrons encountering their next collision within a specified time frame.

The formula for the exponential distribution is given by \(n(t) dt = \frac{N e^{-t/\tau}}{\tau} dt\), where:
  • \(n(t) dt\) is the number of electrons colliding between time \(t\) and \(t+dt\).
  • \(N\) is the total number of electrons.
  • \(\tau\) is the mean free time, or average time between collisions.
  • \(e^{-t/\tau}\) is the exponential decay function describing how the probability decreases over time.

One of the essential properties of the exponential distribution is that the probability decreases exponentially with time, which means that most events (electron collisions, in this case) happen early. The distribution gives us insight into how frequently electrons interact with lattice ions in a given sample.
Mean Free Time
Mean free time, \(\tau\), is an important concept in collision theory, representing the average time interval between successive collisions. For electrons, this is the average duration between successive collisions with ions or other scattering sites in a material. The mean free time can be derived from the experimental data of electron collisions and is a key parameter in understanding the movement of electrons in a solid material.

In equations like \(\int_{0}^{\infty} n(t) dt = N\), mean free time helps establish the scale for the exponential decay of electron collisions, as seen in the exponential distribution mentioned earlier.

The mean time of collision, denoted as \(\bar{t}\), turns out to be equal to the mean free time \(\tau\), providing a useful simplification. It represents not just an average duration, but a fundamental property of the material or system being studied, allowing scientists and engineers to predict behavior in similar conditions.
Collision Theory
Collision theory deals with how and why reactions occur, focusing on reactant particles' collisions. In statistical mechanics, collision theory helps model the interactions of particles such as electrons within atoms or between different materials.

To describe electron behavior, collision theory assumes that all particles must collide with sufficient energy to create a reaction, a central idea when looking at conductivity and resistance in materials.

This exercise's key mathematical derivations connect to collision theory by predicting how many electrons will collide and when, using statistical methods. Substituting and simplifying integrals in the solution gives insights into collision rates and average times, aligning closely with collision theory principles.
Electron Scattering
Electron scattering is a process where electrons deviate from their original path due to collisions with atoms or lattice ions. It is a fundamental phenomenon in statistical mechanics and is significant in understanding electrical conduction and resistance in materials.

When electrons travel through a solid, they encounter ions that cause changes in their trajectory through scattering. This loss of directionality increases resistance and heats the material.

The mathematical modeling of these collision events, as seen in exponential distributions, mean free time, and derived equations, helps quantify and predict electron behavior in various materials. Solving problems related to electron scattering empowers scientists to design materials with desired electrical properties by understanding and controlling scattering rates.

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Most popular questions from this chapter

(a) Calculate the ionic cohesive energy for \(\mathrm{KCA}\) which has the same crystal structure as NaCI. Take \(r_{0}=\) (b) Calculate the atomic cohesive \(0.314 \mathrm{~nm}\) and \(m=9\). energy of KCA by using the facts that the ionization cnergy of potassium is \(4.34 \mathrm{eV}\) (that is, \(\mathrm{K}+4.34 \mathrm{eV} \rightarrow\) \(\mathrm{K}^{+}+e\) ) and that the electron affinity of chlorine is \(3.61 \mathrm{eV}\left(\mathrm{Cl}^{-}+3.61 \mathrm{eV} \rightarrow \mathrm{Cl}+e\right)\)

The contribution of a single electron or hole to the electric conductivity of a semiconductor can be expressed by an important property called the mobility. The mobility, \(\mu\), is defined as the particle drift speed per unit electric field or, in terms of a formula, \(\mu=\tau_{\mathrm{d}} / E\) Note that mobility describes the ease with which charge carriers drift in an electric field and that the mobility of an electron, \(\mu_{x b}\), may be different from the mobility of a hole, \(\mu_{p}\) in the same material. (a) Show that the current density may be written in terms of mobility as \(J=\operatorname{ne\mu}_{n} E\), where \(\mu_{n}=e \tau / m_{\mathrm{c}}-\) (b) Show that when both electrons and holes are present, the conductivity may be expressed in terms of \(\mu_{p}\) and \(\mu_{n}\) as \(\sigma=n e \mu_{n}+p r \mu_{p}\), where \(n\) is the electron concentration and \(p\) is the hole concentration. (c) If a germanium sample has \(\mu_{n}=3900 \mathrm{~cm}^{2} / \mathrm{V} \cdot \mathrm{s}\), calculate the drift speed of an electron when a field of \(100 \mathrm{~V} / \mathrm{cm}\) is applied. (d) A pure (intrinsic) sample of germanium has \(\mu_{n}=3900 \mathrm{~cm}^{2} / \mathrm{V} \cdot \mathrm{s}\) and \(\mu_{\hat{p}}=1900 \mathrm{~cm}^{2} / \mathrm{V} \cdot \mathrm{s}\). If the hole concentration is equal to \(3.0 \times 10^{13} \mathrm{~cm}^{-3}\), calculate the resistivity of the sample.

Determine the current generated in a superconducting ring of niobium metal \(2.00 \mathrm{~cm}\) in diameter when a \(0.0200\) -T magnetic field directed perpendicular to the ring is suddenly decreased to zero. The inductance of the ring is \(3.10 \times 10^{-8} \mathrm{H}\).

Under pressure, liquid helium can solidify as each atom bonds with four others, and each bond has an average energy of \(1.74 \times 10^{-23} \mathrm{~J}\). Find the latent heat of fusion for helium in joules per gram. (The molar mass of He is \(4.00 \mathrm{~g} / \mathrm{mol}\).)

The energy gap for \(\mathrm{Si}\) at \(300 \mathrm{~K}\) is \(1.14 \mathrm{eV}\). (a) Find the lowest-frequency photon that will promote an electron from the valence band to the conduction band of silicon. (b) What is the wavelength of this photon?
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