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USAPhO 2011, Полуфинал — A4

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Всички материали от USAPhO 2011

AAPT UNITED STATES PHYSICS TEAM AIP 2011 — Semifinal Exam · 25 т.

Внимание: Бележка към темата

Printed source errors are preserved: A1’s solution labels both quoted volumes V₀ and says “we used the fat that”; A2’s solution reverses its graph-variable description; A4 prints ω = f/(2π). The solutions repeat the question statements and common instructions; they are shown once here. No individual subpart point values are printed.

Important Instructions for the Exam Supervisor

DO NOT DISTRIBUTE THIS PAGE

Important Instructions for the Exam Supervisor

This examination consists of two parts.
Part A has four questions and is allowed 90 minutes.
Part B has two questions and is allowed 90 minutes.
The first page that follows is a cover sheet. Examinees may keep the cover sheet for both parts of the exam.
The parts are then identified by the center header on each page. Examinees are only allowed to do one part at a time, and may not work on other parts, even if they have time remaining.
Allow 90 minutes to complete Part A. Do not let students look at Part B. Collect the answers to Part A before allowing the examinee to begin Part B. Examinees are allowed a 10 to 15 minutes break between parts A and B.
Allow 90 minutes to complete Part B. Do not let students go back to Part A.
Ideally the test supervisor will divide the question paper into 3 parts: the cover sheet (page 2), Part A (pages 3-4), and Part B (pages 6-7). Examinees should be provided parts A and B individually, although they may keep the cover sheet.
The supervisor must collect all examination questions, including the cover sheet, at the end of the exam, as well as any scratch paper used by the examinees. Examinees may not take the exam questions. The examination questions may be returned to the students after March 31, 2011.
Examinees are allowed calculators, but they may not use symbolic math, programming, or graphic features of these calculators. Calculators may not be shared and their memory must be cleared of data and programs. Cell phones, PDA’s or cameras may not be used during the exam or while the exam papers are present. Examinees may not use any tables, books, or collections of formulas.
Please provide the examinees with graph paper for Part A.

Указания

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

Work Part A first. You have 90 minutes to complete all four problems. Each question is worth 25 points. Do not look at Part B during this time.
After you have completed Part A you may take a break.
Then work Part B. You have 90 minutes to complete both problems. Each question is worth 50 points. Do not look at Part A during this time.
Show all your work. Partial credit will be given. Do not write on the back of any page. Do not write anything that you wish graded on the question sheets.
Start each question on a new sheet of paper. Put your AAPT ID number, your name, the question number and the page number/total pages for this problem, in the upper right hand corner of each page. For example,

AAPT ID #

Doe, Jamie

A1 - 1/3
A hand-held calculator may be used. Its memory must be cleared of data and programs. You may use only the basic functions found on a simple scientific calculator. Calculators may not be shared. Cell phones, PDA’s or cameras may not be used during the exam or while the exam papers are present. You may not use any tables, books, or collections of formulas.
Questions with the same point value are not necessarily of the same difficulty.
In order to maintain exam security, do not communicate any information about the questions (or their answers/solutions) on this contest until after April 1, 2011.

Possibly Useful Information

Possibly Useful Information. You may use this sheet for both parts of the exam.

$g = 9.8\ \mathrm{N/kg}$	$G = 6.67 \times 10^{-11}\ \mathrm{N \cdot m^2/kg^2}$
$k = 1/4\pi\epsilon_0 = 8.99 \times 10^{9}\ \mathrm{N \cdot m^2/C^2}$	$k_{\mathrm{m}} = \mu_0/4\pi = 10^{-7}\ \mathrm{T \cdot m/A}$
$c = 3.00 \times 10^{8}\ \mathrm{m/s}$	$k_{\mathrm{B}} = 1.38 \times 10^{-23}\ \mathrm{J/K}$
$N_{\mathrm{A}} = 6.02 \times 10^{23}\ (\mathrm{mol})^{-1}$	$R = N_{\mathrm{A}} k_{\mathrm{B}} = 8.31\ \mathrm{J/(mol \cdot K)}$
$\sigma = 5.67 \times 10^{-8}\ \mathrm{J/(s \cdot m^2 \cdot K^4)}$	$e = 1.602 \times 10^{-19}\ \mathrm{C}$
$1\mathrm{eV} = 1.602 \times 10^{-19}\ \mathrm{J}$	$h = 6.63 \times 10^{-34}\ \mathrm{J \cdot s} = 4.14 \times 10^{-15}\ \mathrm{eV \cdot s}$
$m_e = 9.109 \times 10^{-31}\ \mathrm{kg} = 0.511\ \mathrm{MeV/c^2}$	$(1+x)^n \approx 1 + nx$ for $\lvert x\rvert \ll 1$
$\sin\theta \approx \theta - \frac{1}{6}\theta^3$ for $\lvert\theta\rvert \ll 1$	$\cos\theta \approx 1 - \frac{1}{2}\theta^2$ for $\lvert\theta\rvert \ll 1$

Условие

In this problem we consider a simplified model of the electromagnetic radiation inside a cubical box of side length $L$ . In this model, the electric field has spatial dependence

E(x, y, z) = E_0 \sin(k_x x)\sin(k_y y)\sin(k_z z)

where one corner of the box lies at the origin and the box is aligned with the $x$ , $y$ , and $z$ axes. Let $h$ be Planck’s constant, $k_B$ be Boltzmann’s constant, and $c$ be the speed of light.

a. The electric field must be zero everywhere at the sides of the box. What condition does this impose on $k_x$ , $k_y$ , and $k_z$ ? (Assume that any of these may be negative, and include cases where one or more of the $k_i$ is zero, even though this causes $E$ to be zero.)

b. In the model, each permitted value of the triple $(k_x, k_y, k_z)$ corresponds to a quantum state. These states can be visualized in a state space, which is a notional three-dimensional space with axes corresponding to $k_x$ , $k_y$ , and $k_z$ . How many states occupy a volume $s$ of state space, if $s$ is large enough that the discreteness of the states can be ignored?

c. Each quantum state, in turn, may be occupied by photons with frequency $\omega = \frac{f}{2\pi} = c|\mathbf{k}|$ , where

|\mathbf{k}| = \sqrt{{k_x}^2 + {k_y}^2 + {k_z}^2}

In the model, if the temperature inside the box is $T$ , no photon may have energy greater than $k_B T$ . What is the shape of the region in state space corresponding to occupied states?

d. As a final approximation, assume that each occupied state contains exactly one photon. What is the total energy of the photons in the box, in terms of $h$ , $k_B$ , $c$ , $T$ , and the volume of the box $V$ ? Again, assume that the temperature is high enough that there are a very large number of occupied states. (Hint: divide state space into thin regions corresponding to photons of the same energy.)

Note that while many details of this model are extremely inaccurate, the final result is correct except for a numerical factor.

STOP: Do Not Continue to Part B

If there is still time remaining for Part A, you should review your work for Part A, but do not continue to Part B until instructed by your exam supervisor.

Отговори

Решение

a. The boundary conditions require $\sin(k_x L) = 0$ , so that

k_x L = n_x \pi

for any integer $n_x$ , with similar conditions for $k_y$ and $k_z$ .

b. In the abstract state space, the states are spaced a distance $\pi/L$ apart. Each can therefore be thought of as occupying volume $\pi^3/L^3$ , and the number of states in the volume $s$ is

N = \frac{L^3 s}{\pi^3}.

c. A photon’s energy is $E = \hbar\omega = \hbar c|\mathbf{k}|$ , where $\hbar = h/2\pi$ . Thus the occupied states obey

\hbar c|\mathbf{k}| \le k_B T.

This corresponds to a ball of radius $k_{\max} = k_B T/\hbar c$ in state space centered at the origin.

d. Naively, we would have to perform a triple integral over state space. However, the energy of a photon depends only on its distance $|\mathbf{k}|$ from the origin in state space. Hence we can integrate over spherical shells. Consider a shell bounded by radii $k$ and $k + dk$ . The volume of this region is

ds = 4\pi k^2 dk.

Each state in this region contains a single photon with energy $\hbar ck$ , so the shell yields energy

dE = \hbar ck\,\frac{L^3}{\pi^3}\,ds = \frac{4}{\pi^2}\hbar c L^3 k^3\,dk.

From our work above, $k$ ranges from zero to $k_{\max}$ , so the total energy is

E = \frac{4}{\pi^2}\hbar c L^3 \int_0^{k_{\max}} k^3\,dk = \frac{\hbar c L^3}{\pi^2} k_{\max}^4.

Substituting $V = L^3$ and $h = 2\pi\hbar$ , this simplifies to

E = \frac{8\pi k_B^4}{h^3 c^3} V T^4.

Note that everything has come out right, including the $T^4$ factor seen in the Stefan-Boltzmann law, though the numerical prefactor is wrong.

Оригинал в Архива: USAPhO-2011.pdf · официални решения: USAPhO-2011-Solutions.pdf