Съдържание

Fundamental constants Astronomical data Approximations Gauss’s formulae The Sun The Earth and Moon Условие Решение

IOAA 2023, Теория — Задача 8. Neutrinos

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Всички материали от IOAA 2023

IOAA 2023 — Theoretical competition · 20 т.

Указания

Do not touch envelopes until the start of the examination.
The theoretical examination lasts for 5 hours and is worth a total of 250 marks.
There are Answer Sheets for carrying out detailed work and Working Sheets for rough work, which are already marked with your student code and question number.
Use only the answer sheets for a particular question for your answer. Please write only on the printed side of the sheet. Do not use the reverse side. If you have written something on any sheet which you do not want to be evaluated, cross it out.
Use as many mathematical expressions as you think may help the evaluator to better understand your solutions. The evaluator may not understand your language. If it is necessary to explain something in words, please use short phrases (if possible in English).
You are not allowed to leave your work desk without permission. If you need any assistance (malfunctioning calculator, need to visit a restroom, etc.), please draw the attention of the supervisor.
The beginning and end of the examination will be indicated by the supervisor. The remaining time will be displayed on a clock.
At the end of the examination you must stop writing immediately. Put everything back in the envelope and leave it on the table.
Once all envelopes are collected, your student guide will escort you out of the examination room.
A list of constants and useful relations are included in the envelope.

Table of Constants

Fundamental constants

Constant	Symbol	Value
Speed of light in vacuum	$c$	$2.998\times10^8\,\mathrm{m\,s^{-1}}$
Planck constant	$h$	$6.626\times10^{-34}\,\mathrm{J\,s}$
Boltzmann constant	$k_B$	$1.381\times10^{-23}\,\mathrm{J\,K^{-1}}$
Stefan-Boltzmann constant	$\sigma$	$5.670\times10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}$
Elementary charge	$e$	$1.602\times10^{-19}\,\mathrm C$
Universal gravitational constant	$G$	$6.674\times10^{-11}\,\mathrm{N\,m^2\,kg^{-2}}$
Universal electric constant	$\epsilon_0$	$8.854\times10^{-12}\,\mathrm{m^{-3}\,kg^{-1}\,s^4\,A^2}$
Universal gas constant	$R$	$8.315\,\mathrm{J\,mol^{-1}\,K^{-1}}$
Avogadro constant	$N_A$	$6.022\times10^{23}\,\mathrm{mol^{-1}}$
Wien’s displacement constant	$b=\lambda_mT$	$2.898\times10^{-3}\,\mathrm{m\,K}$
Mass of electron	$m_e$	$9.109\times10^{-31}\,\mathrm{kg}$
Mass of proton	$m_p$	$1.673\times10^{-27}\,\mathrm{kg}$
Mass of neutron	$m_n$	$1.675\times10^{-27}\,\mathrm{kg}$
Mass of Helium nucleus	$m_{\mathrm{He}}$	$6.645\times10^{-27}\,\mathrm{kg}$
Atomic mass unit (a.m.u., Dalton)		$1.661\times10^{-27}\,\mathrm{kg}$

Astronomical data

Data	Symbol	Value
Hubble constant	$H_0$	$70\,\mathrm{km\,s^{-1}\,Mpc^{-1}}$
North Ecliptic Pole ( $J2000.0$ )	$(\alpha_E,\delta_E)$	$(18\mathrm h\,00\mathrm m\,00\mathrm s,+66^\circ33'39'')$
North Galactic Pole ( $J2000.0$ )	$(\alpha_G,\delta_G)$	$(12\mathrm h\,51\mathrm m\,26\mathrm s,+27^\circ07'42'')$
1 jansky	$1\,\mathrm{Jy}$	$10^{-26}\,\mathrm{W\,m^{-2}\,Hz^{-1}}$
1 parsec	$1\,\mathrm{pc}$	$3.086\times10^{16}\,\mathrm m=206\,265\,\mathrm{au}=3.262\,\mathrm{ly}$
1 astronomical unit (au)	$1\,\mathrm{au}$	$1.496\times10^{11}\,\mathrm m$
1 sidereal day	$T_{\mathrm{SD}}$	$23.93444\,\mathrm h=23\mathrm h\,56\mathrm m\,04\mathrm s$
1 tropical year		$365.2422$ solar days
1 sidereal year		$365.2564$ solar days

Approximations and Gauss’s formulae

Approximations

(1+x)^n\approx1+nx

(1+x)(1+y)\approx1+x+y\quad\text{if }x\ll1\text{ and }y\ll1

Gauss’s formulae

Spherical law of cosines: $\cos a=\cos b\cos c+\sin b\sin c\cos A$

Spherical law of sines: $\dfrac{\sin A}{\sin a}=\dfrac{\sin B}{\sin b}=\dfrac{\sin C}{\sin c}$ .

The Sun, Earth and Moon

The Sun

Data	Symbol	Value
Solar luminosity	$L_\odot$	$3.826\times10^{26}\,\mathrm W$
Apparent angular diameter of Sun	$\theta_\odot$	$32'$
Effective temperature of Sun	$T_{\mathrm{eff},\odot}$	$5778\,\mathrm K$
Apparent visual magnitude		$-26.75$
Absolute visual magnitude		$+4.82$
Apparent bolometric magnitude		$-26.83$
Absolute bolometric magnitude		$+4.74$
Distance of the Sun from the Galactic centre		$\approx8\,\mathrm{kpc}$

The Earth and Moon

Data	Symbol	Value
Obliquity of the ecliptic (Earth)	$\epsilon$	$23.5'$
Platonic year (period of precession of Earth’s axis)		$25\,765$ years
Apparent visual magnitude of full Moon		$-12.74$
Apparent angular diameter of Moon	$\theta_L$	$31'$
Inclination of the lunar orbit to the ecliptic		$05^\circ08'43''$
Inclination of the lunar equator to its orbital plane		$6.687^\circ$
Lunar sidereal month	$T_{\mathrm{SL}}$	$27.321661\,\mathrm d=655.71986\,\mathrm h$
Synodic month		$29.530589\,\mathrm d$
Tropical month		$27.321582\,\mathrm d$
Anomalistic month (the average time between two successive Moon’s perigees)		$27.554550\,\mathrm d$
Draconic month (time between two successive transits of the Moon through the same node)		$27.212221\,\mathrm d$

The Solar System

Object	Mean radius [km]	Mass [kg]	Semimajor axis [au]	Eccentricity
Sun	$695\,700$	$1.988\times10^{30}$	—	—
Mercury	$2\,440$	$3.301\times10^{23}$	$0.387$	$0.206$
Venus	$6\,052$	$4.867\times10^{24}$	$0.723$	$0.007$
Earth	$6\,378$	$5.972\times10^{24}$	$1.000$	$0.016\,710$
Moon	$1\,737$	$7.346\times10^{22}$	$3.844\times10^5\,\mathrm{km}$	$0.054\,900$ (range $0.026$ – $0.077$ )
Mars	$3\,390$	$6.417\times10^{23}$	$1.524$	$0.093$
Jupiter	$69\,911$	$1.898\times10^{27}$	$5.203$	$0.048$
Saturn	$58\,232$	$5.683\times10^{26}$	$9.537$	$0.054$
Uranus	$25\,362$	$8.681\times10^{25}$	$19.189$	$0.047$
Neptune	$24\,622$	$1.024\times10^{26}$	$30.070$	$0.009$

General Marking Scheme

Case	Marking
Using incorrect physical concept (despite correct answers)	No points given
Giving correct answer without detailed calculation	Deduct 50% of the marks for that part
Minor mistakes in the calculations, e.g., wrong signs, symbols, substitutions	Deduct 20% of the marks for that part
Units missing from final answers	Deduct 0.5 pts
Too few or too many significant figures in the final answer	Deduct 0.5 pts
Error resulting from another error in an earlier part for which the student already lost marks, if the answer is physically reasonable.	Full points (i.e., no deductions)
Error resulting from another error in an earlier part, where the student should have realised the answer was physically unreasonable.	Deduct 20% of the marks for that part

For example, if due to an error in an earlier part, the student calculates the mass of a star as $2.5\times10^{30}\,\mathrm{kg}$ instead of $2\times10^{30}\,\mathrm{kg}$ , they will only lose marks for the earlier part. However, if, for the same reason, a student calculates the mass as $2\times10^{25}\,\mathrm{kg}$ , they should realize this is wrong (a few times the Earth’s mass) and thus should lose some marks for this part as well.

Условие

In a simplified model of a supernova explosion, the core of a star, composed of pure iron $^{56}_{26}\mathrm{Fe}$ nuclei with a total mass of $1\,M_\odot$ , changes into a neutron star composed of individual electrons, protons and neutrons in numerical proportions of 1:1:8. This process is called ‘neutronization’ and results in the emission of a large number of neutrinos.

Calculate the solar neutrino flux on Earth. How much larger would the flux of neutrinos reaching the Earth from the supernova be than the steady neutrino emission of the Sun, if the supernova exploded in the centre of the Galaxy and the process of neutronization of the core took about $0.01\,\mathrm{s}$ ? Give an order-of-magnitude answer.

Решение

Neutrino emission of the supernova:

From the table of constants, the core mass $M=1M_\odot=1.988\cdot10^{30}\,\mathrm{kg}$ .

Mass of one atom of iron $^{56}_{26}\mathrm{Fe}$ :

m_{\mathrm{Fe}}=56\,\mathrm{Da}=56\times1.661\times10^{-27}\,\mathrm{kg}=9.3016\times10^{-26}\,\mathrm{kg}.

Note: atomic mass is expressed in atomic mass units symbol a.m.u. or u, also called Daltons, symbol Da. Students may use any of these notations in their solutions.

The number of $^{56}_{26}\mathrm{Fe}$ nuclei in the core is therefore:

n_{\mathrm{Fe}}=M/m_{\mathrm{Fe}}=2.1373\times10^{55}

(1 points)

and the initial number of nucleons is:

n_{\mathrm{nuc}}=56n_{\mathrm{Fe}}=1.1969\times10^{57}.

(1 points)

Correspondingly, the initial number of protons is $n_{\mathrm{p}}=26n_{\mathrm{Fe}}$ and the initial number of neutrons is $n_{\mathrm{n}}=30n_{\mathrm{Fe}}$ . (2 points)

We assume all the nucleons in the original $^{56}_{26}\mathrm{Fe}$ nuclei are converted to individual nucleons in the given proportions (1 proton : 8 neutrons). Therefore after the explosion,

n'_{\mathrm{n}}=\frac89 n_{\mathrm{nuc}}=1.0634\times10^{57}

neutrons and

n'_{\mathrm{p}}=\frac19 n_{\mathrm{nuc}}=1.3299\times10^{56}

protons remain, and thus

n_{\mathrm{p}}-n'_{\mathrm{p}}=26n_{\mathrm{Fe}}-n'_{\mathrm{p}}=4.2271\times10^{56}

(4 points)

protons are changed. Since 1 neutrino is produced for every converted proton ( $\mathrm{p}+\mathrm{e}^-\rightarrow\mathrm{n}+\nu_e$ ), the neutrino flux is:

I_\nu=\frac{4.2271\times10^{56}}{0.01\,\mathrm{s}}=4.2271\times10^{58}\,\mathrm{neutrinos\ s^{-1}}.

(2 points)

The flux density observed at Earth is given by:

F_{\mathrm{SN}}=\frac{I_\nu}{4\pi d_{\mathrm{Gal}}^2}.

Substituting $d_{\mathrm{Gal}}=8\,\mathrm{kpc}=8000\times3.086\times10^{16}\,\mathrm{m}=2.4688\times10^{20}\,\mathrm{m}$ ,

F_{\mathrm{SN}}\approx5.5\times10^{16}\,\mathrm{neutrinos\ s^{-1}\,m^{-2}}.

(2 points)

Neutrino flux from the Sun:

We can assume that the primary source of Solar luminosity is the p–p reaction,

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

Neglecting the electrons and neutrinos, the change in mass $\Delta m$ is:

\Delta m=4\cdot1.673\times10^{-27}\,\mathrm{kg}-6.645\times10^{-27}\,\mathrm{kg}=4.7\times10^{-29}\,\mathrm{kg}.

and thus from $E=(\Delta m)c^2$ the amount of energy released is $\approx4.2244\times10^{-12}\,\mathrm{J}$ .

Solar luminosity is $L_\odot=3.826\times10^{26}\,\mathrm{W}$ , so the number of reactions per second is approximately:

\frac{3.826\times10^{26}}{4.2244\times10^{-12}}=9.057\times10^{37}.

With 2 neutrinos released per reaction, this gives a Solar neutrino flux of:

I_{\nu\odot}\approx1.8\times10^{38}\,\mathrm{neutrinos\ s^{-1}}.

(If the student remembers that the p–p reaction releases $26.73\,\mathrm{MeV}=4.2826\times10^{-12}\,\mathrm{J}$ , they will get $8.9338\times10^{37}$ reactions per second and the same final answer.)

(getting to $I_{\nu\odot}$ : 5 points)

At the distance of Earth, the observed flux density is given by:

F_\odot=\frac{I_{\nu\odot}}{4\pi d_\odot^2},

where $d_\odot=1\,\mathrm{au}=1.496\times10^{11}\,\mathrm{m}$ , and so:

F_\odot\approx6.4\times10^{14}\,\mathrm{neutrinos\ s^{-1}\,m^{-2}}.

(1 point)

The final ratio is therefore:

F_{SN}/F_\odot\approx\frac{5.5\times10^{16}}{6.4\times10^{14}}\approx86\approx100,

or two orders of magnitude. (2 points)

Original PDF comments

The original solution PDF contains two comments attributed to IOAA on this page:

“In previuos erratic version it was "... - 1".”

“In previous erratic version it was L_2/L_1 = ...”

Бележки за изданието

The English question packet includes a Ukrainian answer form for problem 9. The constants table prints Earth’s obliquity as 23.5′ and the Moon’s semimajor axis in km under an au heading. In the eclipse key, the magnitude addition and luminosity-ratio labels differ from its derivation and results; the original PDF correction comments are retained. Aldebaran’s key prints 4412 km/h and then uses 4416 in the angular-speed numerator. The X-ray question and key allocate points to parts (c) and (d) differently. DART’s alternative calculation prints both 0.17622 and 0.17662 m/s. The LISA key labels the component-mass calculation Part (b), repeats one algebra line and prints the period factor with a positive square. These printed discrepancies remain literal.

Оригинал в Архива: theory.pdf · официални решения: th_solutions.pdf