Съдържание

Условие Решение

USAPhO 2001, Полуфинал — A4

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от USAPhO 2001

2001 Semi-Final Exam · 25 т.

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.
After you have completed Part A, you may take a break.
Then work Part B. You have 90 minutes to complete both problems.
Show all your work. Partial credit will be given.
Start each question on a new sheet of paper. Be sure to put your name in the upper right-hand corner of each page, along with the question number and the page number/total pages for this problem. For example,

Doe, Jamie

A1 – 1/3

A hand-held calculator may be used. Its memory must be cleared of data and programs. It may not be used in graphing mode. Calculators may not be shared. You may not use any tables, books, or collections of formulas. You may use a ruler or straight edge.
Questions with the same point value are not necessarily of the same difficulty.
Good luck!

Possibly Useful Information

Quantity	Value
Gravitational field at the Earth's surface	$g=9.8$ N/kg
Newton's gravitational constant	$G=6.67\times10^{-11}$ N·m $^2$ /kg $^2$
Coulomb's constant	$k=1/4\pi\epsilon_0=9.0\times10^9$ N·m $^2$ /C $^2$
Biot-Savart constant	$k_m=\mu_0/4\pi=10^{-7}$ T·m/A
Speed of light in a vacuum	$c=3.0\times10^8$ m/s
Boltzmann's constant	$k_B=1.38\times10^{-23}$ J/K
Avogadro's number	$N_A=6.02\times10^{23}$ (mol) $^{-1}$
Ideal gas constant	$R=N_Ak_B=8.31$ J/(mol·K)
Stefan-Boltzmann constant	$\sigma=5.67\times10^{-8}$ J/(s·m $^2$ ·K $^4$ )
Elementary charge	$e=1.6\times10^{-19}$ C
1 electron volt	$1\text{ eV}=1.6\times10^{-19}$ J
Planck's constant	$h=6.63\times10^{-34}\text{ J}\cdot\text{s}=4.14\times10^{-15}\text{ eV}\cdot\text{s}$
Electron mass	$m=9.1\times10^{-31}\text{ kg}=0.51\text{ MeV}/c^2$
Binomial expansion	$(1+x)^n\simeq1+nx\quad\text{for }\lvert x\rvert\ll1$

Условие

The solar radius is $6.96\times10^8$ m. The sun has a surface temperature of $5.80\times10^3$ K and is $1.50\times10^{11}$ m from the earth. Assume the sun is a perfect absorber/emitter. Suppose this thermal energy comes from the following set of nuclear reactions:

\begin{aligned}{}^1_1\mathrm H+{}^1_1\mathrm H&\rightarrow{}^2_1\mathrm H+e^++\nu\\{}^1_1\mathrm H+{}^1_1\mathrm H&\rightarrow{}^2_1\mathrm H+e^++\nu\\{}^1_1\mathrm H+{}^2_1\mathrm H&\rightarrow{}^3_2\mathrm{He}+\gamma\\{}^1_1\mathrm H+{}^2_1\mathrm H&\rightarrow{}^3_2\mathrm{He}+\gamma\\{}^3_2\mathrm{He}+{}^3_2\mathrm{He}&\rightarrow{}^4_2\mathrm{He}+{}^1_1\mathrm H+{}^1_1\mathrm H\end{aligned}

where $e^+$ represents a positron, $\nu$ represents a neutrino, and $\gamma$ represents a gamma ray. This set of reactions releases a thermal energy of 26.2 MeV. Assume that the sun's power production and temperature at any depth does not change with time. Find the intensity (number per second per square meter) of solar neutrinos at the location of the Earth and the number of solar neutrinos in the adult human body. Assume that the adult human body has a volume of $8.00\times10^{-2}$ m $^3$ and that neutrinos travel at the speed of light.

Решение

The sun's radius is $R_S=6.96\times10^8$ m and its surface temperature is $T=5.80\times10^3$ K. The sun-Earth distance is $R_{ES}=1.50\times10^{11}$ m. Each set of reactions release 2 neutrinos and thermal energy $Q=26.2\text{ MeV}=26.2\text{ MeV}(1.6\times10^{-13}\text{ J/MeV})=4.192\times10^{-12}$ J. The energy/neutrino ratio is $2.096\times10^{-12}\text{ J}/\nu$ . For a perfect absorber emitter, $e=1$ .

The power radiated by the sun is $P=e\sigma AT^4=e\sigma(4\pi R_S^2)T^4$ .

The solar intensity reaching the Earth is

I=\frac P{A_{ES}}=\frac{e\sigma4\pi R_S^2T^4}{4\pi R_{ES}^2}=\frac{(1)(5.67\times10^{-8}\text{ J}/(\text{s}\cdot\text{m}^2\cdot\text{K}^4))(6.96\times10^8\text{ m})^2(5800\text{ K})^4}{(1.50\times10^{11}\text{ m})^2}=1.38\times10^3\frac{\text J}{\text{s}\cdot\text{m}^2}

To find the neutrino intensity $I_\nu$ , divide the solar intensity $I$ by the energy/neutrino ratio.

I_\nu=\frac I{Q/\nu}=\frac{1380\text{ J}/(\text{s}\cdot\text{m}^2)}{2.096\times10^{-12}\text{ J}/\nu}=6.59\times10^{14}\text{ s}^{-1}\text{m}^{-2}

The neutrino intensity is also equal to their velocity $c$ times their number density $N/V$ , $I_\nu=cN/V$ .

Solving for $N$ ,

N=\frac{I_\nu V}c=\frac{(6.59\times10^{14}\text{ s}^{-1}\text{m}^{-2})(8.00\times10^{-2}\text{ m}^3)}{3.0\times10^8\text{ m/s}}=1.76\times10^5.

Оригинал в Архива: Semifinal 2001.pdf · официални решения: Semifinal 2001 Solutions.pdf