Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 36: Problem 55

Consider a system made up of \(N\) particles. The average energy per particle is given by \(\langle E\rangle=\left(\sum E_{i} e^{-E_{i} / k_{B} T}\right) / Z\) where \(Z\) is the partition function defined in equation \(36.29 .\) If this is a two-state system with \(E_{1}=0\) and \(E_{2}=E\) and \(g_{1}=\) \(g_{2}=1,\) calculate the heat capacity of the system, defined as \(N(d\langle E\rangle / d T)\) and approximate its behavior at very high and very low temperatures (that is, \(k_{\mathrm{B}} T \gg 1\) and \(k_{\mathrm{B}} T \ll 1\) ).

Short Answer

Expert verified
Answer: At very high temperatures, the heat capacity is approximately 0, while at very low temperatures, the heat capacity is proportional to \(N E^2 / k_{B} T^2\).

Step by step solution

01

Calculate the partition function Z

Since we have a two-state system with \(E_{1}=0\) and \(E_{2}=E\), and \(g_{1}=g_{2}=1\), the partition function Z is given by the sum of the Boltzmann factors for these two states: $$Z = g_{1}e^{-E_{1}/k_BT} + g_{2}e^{-E_{2}/k_BT} = e^0 + e^{-E/k_BT} = 1 + e^{-E/k_BT}$$
02

Calculate the average energy per particle

Using the formula for average energy per particle, let's plug in the values for \(E_{1}\), \(E_{2}\), and Z: $$\langle E\rangle = \frac{\sum E_{i} e^{-E_{i} / k_{B} T}}{Z} = \frac{0\cdot e^0 + E\cdot e^{-E/k_BT}}{1 + e^{-E/k_BT}} = \frac{E\cdot e^{-E/k_BT}}{1 + e^{-E/k_BT}}$$
03

Differentiate the average energy per particle with respect to temperature T

Now let's differentiate \(\langle E\rangle\) with respect to T, using quotient rule: $$\frac{d\langle E\rangle}{dT} = \frac{-E^{2}/k_{B} e^{-E/k_BT}(1 + e^{-E/k_BT}) - E\cdot e^{-E/k_BT}(-E/k_BT)e^{-E/k_BT}}{(1 + e^{-E/k_BT})^2}$$ After some simplification: $$\frac{d\langle E\rangle}{dT} = \frac{E^{2}/k_{B} e^{-E/k_BT}}{k_{B} T^2 (1 + e^{-E/k_BT})^2}$$
04

Calculate the heat capacity

The heat capacity is defined as \(N(d\langle E\rangle / d T)\). Plug in the result from Step 3: $$C = N\frac{E^2 / k_{B} e^{-E/k_BT}}{k_{B} T^2 (1 + e^{-E/k_BT})^2} = \frac{N E^2 e^{-E/k_BT}}{k_{B} T^2 (1 + e^{-E/k_BT})^2}$$
05

Approximate the heat capacity behavior at very high and very low temperatures

At very high temperatures (\(k_BT \gg 1\)), \(e^{-E/k_BT} \approx 0\), so the heat capacity becomes: $$C \approx \frac{N E^2 \cdot 0}{k_{B} T^2 (1 + 0)^2} = 0$$ At very low temperatures (\(k_BT \ll 1\)), \(e^{-E/k_BT} \approx \infty\), so the heat capacity simplifies to: $$C \approx \frac{N E^2 \cdot \infty}{k_{B} T^2 (\infty)^2} = \frac{N E^2}{k_{B} T^2}$$ So, at very high temperatures, the heat capacity is approximately 0, while at very low temperatures, the heat capacity is proportional to \(N E^2 / k_{B} T^2\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Which of the following has the higher temperature? a) a white-hot object c) a blue-hot object b) a red-hot object

Which one of the following statements is true if the intensity of a light beam is increased while its frequency is kept the same? a) The photons gain higher speeds. b) The energy of the photons is increased. c) The number of photons per unit time is increased. d) The wavelength of the light is increased.

Scintillation detectors for gamma rays transfer the energy of a gamma-ray photon to an electron within a crystal, via the photoelectric effect or Compton scattering. The electron transfers its energy to atoms in the crystal, which re-emit it as a light flash detected by a photomultiplier tube. The charge pulse produced by the photomultiplier tube is proportional to the energy originally deposited in the crystal; this can be measured so an energy spectrum can be displayed. Gamma rays absorbed by the photoelectric effect are recorded as a photopeak in the spectrum, at the full energy of the gammas. The Compton-scattered electrons are also recorded, at a range of lower energies known as the Compton plateau. The highest-energy of these form the Compton edge of the plateau. Gamma-ray photons scattered \(180 .^{\circ}\) by the Compton effect appear as a backscatter peak in the spectrum. For gamma-ray photons of energy \(511 \mathrm{KeV}\) calculate the energies of the Compton edge and the backscatter peak in the spectrum.

You are performing a photoelectric effect experiment. Using a photocathode made of cesium, you first illuminate it with a green laser beam \((\lambda=514.5 \mathrm{nm})\) of power \(100 \mathrm{~mW}\). Next, you double the power of your laser beam, to \(200 \mathrm{~mW}\). How will the energies per electron of the electrons emitted by the cathode compare for the two cases?

The temperature of your skin is approximately \(35.0^{\circ} \mathrm{C}\). a) Assuming that it is a blackbody, what is the peak wavelength of the radiation it emits? b) Assuming a total surface area of \(2.00 \mathrm{~m}^{2}\), what is the total power emitted by your skin? c) Based on your answer in (b), why don't you glow as brightly as a light bulb?
See all solutions

Recommended explanations on Physics Textbooks

Fluids

Read Explanation

Particle Model of Matter

Read Explanation

Turning Points in Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Engineering Physics

Read Explanation

Measurements

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free