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Условие Отговори Решение

USAPhO 2010, Полуфинал — Задача 4. Question A4

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Всички материали от USAPhO 2010

AAPT AIP United States Physics Team 2010 — Semifinal Exam · 25 т.

Important Instructions for the Exam Supervisor

DO NOT DISTRIBUTE THIS PAGE

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, 2010.
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.

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 school 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,

School 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 March 31, 2010.

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

The energy radiated by the Sun is generated primarily by the fusion of hydrogen into helium-4. In stars the size of the Sun, the primary mechanism by which fusion takes place is the proton-proton chain. The chain begins with the following reactions:

2\ \mathrm{p} \rightarrow \mathbf{X}_1 + \mathrm{e}^{+} + \mathbf{X}_2\ \ (0.42\ \mathrm{MeV}) \qquad (A4-1)

\mathrm{p} + \mathbf{X}_1 \rightarrow \mathbf{X}_3 + \gamma\ \ (5.49\ \mathrm{MeV}) \qquad (A4-2)

The amounts listed in parentheses are the total kinetic energy carried by the products, including gamma rays. p is a proton, $\mathrm{e}^{+}$ is a positron, $\gamma$ is a gamma ray, and $\mathbf{X}_1$ , $\mathbf{X}_2$ , and $\mathbf{X}_3$ are particles for you to identify.

The density of electrons in the Sun’s core is sufficient that the positron is annihilated almost immediately, releasing an energy $x$ :

\mathrm{e}^{+} + \mathrm{e}^{-} \rightarrow 2\ \gamma\ \ (x) \qquad (A4-3)

Subsequently, two major processes occur simultaneously. The “pp I branch” is the single reaction

2\ \mathbf{X}_3 \rightarrow {}^{4}\mathrm{He} + 2\ \mathbf{X}_4\ \ (y), \qquad (A4-4)

which releases an energy $y$ . The “pp II branch” consists of three reactions:

\mathbf{X}_3 + {}^{4}\mathrm{He} \rightarrow \mathbf{X}_5 + \gamma \qquad (A4-5)

\mathbf{X}_5 + \mathrm{e}^{-} \rightarrow \mathbf{X}_6 + \mathbf{X}_7\ \ (z) \qquad (A4-6)

\mathbf{X}_6 + \mathbf{X}_4 \rightarrow 2\ {}^{4}\mathrm{He} \qquad (A4-7)

where $z$ is the energy released in step A4-6.

a. Identify $\mathbf{X}_1$ through $\mathbf{X}_7$ . $\mathbf{X}_2$ and $\mathbf{X}_7$ are neutral particles of negligible mass. It is useful to know that the first few elements, in order of atomic number, are H, He, Li, Be, B, C, N, O.

b. The mass of the electron is $0.51\ \mathrm{MeV/c^2}$ , the mass of the proton is $938.27\ \mathrm{MeV/c^2}$ , and the mass of the helium-4 nucleus is $3\,727.38\ \mathrm{MeV/c^2}$ . Find the energy released during the production of one helium-4 nucleus, including the kinetic energy of all products and all energy carried by gamma rays.

c. Find the unknown energies $x$ and $y$ above.

d. Step (A4-6) does not proceed as follows because there is insufficient energy.

\mathbf{X}_5 \rightarrow \mathbf{X}_6 + \mathrm{e}^{+} + \mathbf{X}_7

What constraint does this fact place on $z$ ?

e. In which of the reaction steps is the energy carried by any given product the same every time the step occurs? Assume that the kinetic energy carried in by the reactants in each step is negligible, and that the products are in the ground state.

Отговори

Решение

a. We know that in all nuclear processes, total charge is conserved, lepton number (electrons plus neutrinos minus positrons minus antineutrinos), and baryon number (neutrons plus protons) are conserved. As $X_2$ is a neutral particle of negligible mass, $X_1$ must have charge +1 and contain two baryons. Thus it is ${}^{2}\mathrm{H}$ . $X_2$ is then a neutral particle with lepton number +1 and is an electron neutrino.

Similar reasoning shows that $X_3$ is ${}^{3}\mathrm{He}$ , $X_4$ is a proton (i.e. ${}^{1}\mathrm{H}$ ), $X_5$ is ${}^{7}\mathrm{Be}$ , $X_6$ is ${}^{7}\mathrm{Li}$ , and $X_7$ is an electron neutrino.

b. The overall reaction (not including gamma rays) is

4\ \mathrm{p} + 2\ \mathrm{e}^{-} \rightarrow\ {}^{4}\mathrm{He} + 2\ \nu_e

as can be seen from conservation considerations or by combining the given reactions. The energy released is the difference in mass between the reactants and the products; using the given values, this is 26.72 MeV.

c. $x$ is simply twice the mass of the electron, 1.02 MeV. To compute $y$ , note that we can sum the other known energies to obtain the result from the previous problem:

2(0.42\ \mathrm{MeV} + 5.49\ \mathrm{MeV} + 1.02\ \mathrm{MeV}) + y = 26.72\ \mathrm{MeV}

where the factor of 2 arises because two ${}^{3}\mathrm{He}$ are produced in the course of the combined reaction. Solving, $y = 12.86\ \mathrm{MeV}$ .

d. The forbidden reaction produces an energy of $z - 2(0.51\ \mathrm{MeV})$ , as it differs from the naturally occurring one by the consumption of one fewer electron and the production of an additional positron. Since it is forbidden, $z < 1.02\ \mathrm{MeV}$ .

e. Reactions with two products have a single set of product energies, while those with three or more products produce a spectrum of output energies. Students may quote this result; alternatively, observe that in a two-product reaction the conservation of momentum and of energy give two equations in the two unknowns, fixing their values, whereas there are insufficient equations in the case of three or more products.

The reactions with only two products are (A4-2), (A4-3), (A4-5), (A4-6), and (A4-7).

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.

Оригинал в Архива: 2010_SemiFinal.pdf · официални решения: USAPhO-2010-Solutions.pdf