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USAAAO 2021, II кръг — Long 1

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Всички материали от USAAAO 2021

2021 National Astronomy Competition · 40 т.

Внимание: Бележка към темата

All ten whole original Short/Medium/Long problems, 30 lettered parts, 9 nested roman requests and 7 numbered satellite quantities, all original worked keys and marking notes are retained. Printed scientific discrepancies are preserved in attributed local source notes. Medium 3 original calls both masses m; its key introduces a distinct M for the station and assumes $m\lt M$ . A separate undated common archive constants/reference table is fully included with its own source identity; its use as the exact 2021 provided attachment is not established, and its constants do not replace native exam values. The native cover specifies blank paper and a signed declaration. March 20th is the end of the examination period, not an established single sitting date.

1 Instructions (Please Read Carefully)

The top 5 eligible scorers on the NAC will be invited to represent USA at the next IOAA. In order to qualify for the national team, you must be a high school student with US citizenship or permanent residency.

This exam consists of 3 parts: Short Questions, Medium Questions and Long Questions.

The maximum number of points is 180.

The test must be completed within 2.5 hours (150 minutes).

Please solve each problem on a blank piece of paper and mark the number of the problem at the top of the page. The contestant’s full name in capital letters should appear at the top of each solution page. If the contestant uses scratch papers, those should be labeled with the contestant’s name as well and marked as “scratch paper” at the top of the page. Scratch paper will not be graded. Partial credit will be available given that correct and legible work was displayed in the solution.

This is a written exam. Contestants can only use a scientific or graphing calculator for this exam. A table of physical constants will be provided. Discussing the problems with other people is strictly prohibited in any way until the end of the examination period on March 20th. Receiving any external help during the exam is strictly prohibited. This means that the only allowed items are: a calculator, the provided table of constants, a pencil (or pen), an eraser, blank sheets of papers, and the exam. No books or notes are allowed during the exam. Exam is proctored and recorded. You are expected to have your video on at all times.

After reading the instructions, please make sure to sign, affirming that:

All work on this exam is mine.
I did not receive any external aids besides the materials provided.
I am not allowed to discuss the test with others throughout the period of this examination.
Failure to follow these rules will lead to disqualification from the exam.

Signed: ____________________

Common archive reference — Table of Physical Constants

This separate common USAAAO archive reference is undated in the catalogue. Its use as the exact 2021 exam attachment is not established.

Quantity	Symbol	Value
Electron rest mass	$m_e$	$9.109\times10^{-31}\ \mathrm{kg}$
Proton rest mass	$M_p$	$1.6726\times10^{-27}\ \mathrm{kg}$
Electronic charge	$e$	$1.6022\times10^{-19}\ \mathrm C$
Speed of light in free space	$c$	$2.9979\times10^8\ \mathrm{m\ s^{-1}}$
Permeability of free space	$\mu_0$	$4\pi\times10^{-7}\ \mathrm{H\ m^{-1}}$
Permittivity of free space	$\epsilon_0$	$8.854\times10^{-12}\ \mathrm{F\ m^{-1}}$
Planck’s constant	$h$	$6.626\times10^{-34}\ \mathrm{J\ s}$
Reduced Planck’s constant	$\hbar=h/2\pi$	$1.0546\times10^{-34}\ \mathrm{J\ s}$
	$\hbar c$	$197.33\ \mathrm{MeV\ fm}$
Boltzmann’s constant	$k_B$	$1.3807\times10^{-23}\ \mathrm{J\ K^{-1}}$
Gas constant	$\mathcal R=k_B/m_H$	$8.250\times10^3\ \mathrm{J\ kg^{-1}\ K^{-1}}$
Molar gas constant	$R$	$8.315\ \mathrm{J\ mol^{-1}\ K^{-1}}$
Avogadro’s number	$N_A$	$6.022\times10^{23}\ \mathrm{mol^{-1}}$
Standard molar volume		$22.414\times10^{-3}\ \mathrm{m^3\ mol^{-1}}$
Unified atomic mass unit ( $^{12}\mathrm C$ scale)	$u$	$931.5\ \mathrm{MeV}/c^2=1.660538\times10^{-27}\ \mathrm{kg}$
Mass of hydrogen atom	$m_H$	$1.0078u=1.6735\times10^{-27}\ \mathrm{kg}$
Bohr magneton	$\mu_B$	$9.274\times10^{-24}\ \mathrm{A\ m^2}$ or $\mathrm{J\ T^{-1}}$
Nuclear magneton	$\mu_N$	$5.051\times10^{-27}\ \mathrm{A\ m^2}$ or $\mathrm{J\ T^{-1}}$
Proton magnetic moment	$\mu_p$	$2.7928\mu_N$
Neutron magnetic moment	$\mu_n$	$-1.9130\mu_N$
Bohr radius	$a_0$	$5.292\times10^{-11}\ \mathrm m$
Fine structure constant	$\alpha=e^2/(4\pi\epsilon_0\hbar c)$	$(137.04)^{-1}$
Compton wavelength of electron	$\lambda_C=h/(m_ec)$	$2.4263\times10^{-12}\ \mathrm m$
Rydberg’s constant	$R_{\infty}$	$1.0974\times10^7\ \mathrm{m^{-1}}$
	$R_{\infty}hc$	$13.606\ \mathrm{eV}$
Stefan-Boltzmann constant	$\sigma$	$5.671\times10^{-8}\ \mathrm{W\ m^{-2}\ K^{-4}}$
Radiation density constant	$a=4\sigma/c$	$7.561\times10^{-16}\ \mathrm{J\ m^{-3}\ K^{-4}}$
Gravitational constant	$G$	$6.673\times10^{-11}\ \mathrm{N\ m^2\ kg^{-2}}$

Rest masses of some leptons and hadrons in $\mathrm{MeV}/c^2$ :

$e^{\pm}$ 0.5110, $\mu^{\pm}$ 105.66, $\tau^{\pm}$ 1777, $\pi^0$ 134.98, $\pi^{\pm}$ 139.57, $\mathrm K^{\pm}$ 493.7, $\mathrm K^0$ 497.7, $\eta$ 547, $\mathrm D^0$ 1865, $\mathrm D^{\pm}$ 1869, p 938.3, n 939.6, $\Lambda^0$ 1115.7, $\Sigma^+$ 1189, $\Sigma^0$ 1193, $\Sigma^-$ 1197, $\Xi^0$ 1315, $\Xi^-$ 1321, $\Omega^-$ 1672, $\mathrm Z^0$ $91.187\times10^3$ , $\mathrm W^{\pm}$ $80.41\times10^3$ .

Quark	Charge	$I_3$	$S$	$C$	$B$	$T$	Mass ( $\mathrm{GeV}/c^2$ )
u	$+\frac23$	$\frac12$	0	0	0	0	$\sim0.003$
d	$-\frac13$	$-\frac12$	0	0	0	0	$\sim0.006$
c	$+\frac23$	0	0	+1	0	0	$\sim1.25$
s	$-\frac13$	0	-1	0	0	0	$\sim0.11$
t	$+\frac23$	0	0	0	0	+1	174.3
b	$-\frac13$	0	0	0	-1	0	4.2

Common archive reference — Astrophysical Data and identities

Astrophysical Data

Quantity	Symbol	Value
1 astronomical unit	AU	$1.496\times10^{11}\ \mathrm m$
1 parsec	pc	$3.086\times10^{16}\ \mathrm m$
Luminosity of Sun	$L_{\odot}$	$3.85\times10^{26}\ \mathrm W$
Mass of Sun	$M_{\odot}$	$1.989\times10^{30}\ \mathrm{kg}$
Radius of Sun	$R_{\odot}$	$6.96\times10^8\ \mathrm m$
Mass of Earth	$M_E$	$5.9742\times10^{24}\ \mathrm{kg}$
Radius of Earth	$R_E$	$6.3781\times10^6\ \mathrm m$

Other data and conversion factors

Quantity	Symbol	Value
1 ångstrom	Å	$10^{-10}\ \mathrm m$
1 fermi	fm	$10^{-15}\ \mathrm m$
1 barn	b	$10^{-28}\ \mathrm{m^2}$
1 pascal	Pa	$1\ \mathrm{Nm^{-2}}$
1 standard atmosphere		$1.0132\times10^5\ \mathrm{Pa}$
Standard acceleration due to gravity	$g$	$9.807\ \mathrm{m\ s^{-2}}$
1 electron volt	eV	$1.6022\times10^{-19}\ \mathrm J$
	$\mathrm{eV}/hc$	$8.065\times10^5\ \mathrm{m^{-1}}$
	$\mathrm{eV}/k_B$	$1.1604\times10^4\ \mathrm K$
Wavelength of 1 eV photon		$1.2399\times10^{-6}\ \mathrm m$

Trigonometrical identities

\sin(\theta+\phi)=\sin(\theta)\cos(\phi)+\cos(\theta)\sin(\phi)

\cos(\theta+\phi)=\cos(\theta)\cos(\phi)-\sin(\theta)\sin(\phi)

\sin\alpha+\sin\beta=2\sin\frac12(\alpha+\beta)\cos\frac12(\alpha-\beta)

\cos\alpha+\cos\beta=2\cos\frac12(\alpha+\beta)\cos\frac12(\alpha-\beta)

\cos\alpha-\cos\beta=2\sin\frac12(\alpha+\beta)\sin\frac12(\beta-\alpha)

In a triangle ABC, $a/\sin A=b/\sin B=c/\sin C$

and $a^2=b^2+c^2-2bc\cos A$

Prefixes

Prefix	Name	Factor
T	tera	$10^{12}$
G	giga	$10^9$
M	mega	$10^6$
k	kilo	$10^3$
c	centi	$10^{-2}$
m	milli	$10^{-3}$
$\mu$	micro	$10^{-6}$
n	nano	$10^{-9}$
p	pico	$10^{-12}$
f	femto	$10^{-15}$

Условие

4 Long Questions

In the very early universe, everything is in thermodynamic equilibrium and particles are freely created, destroyed, and converted between each other due to the high temperature. In one such process, the reaction converting between neutrons and protons happens at a very high rate. In thermal equilibrium, the relative number density of particle species is given approximately by the Boltzmann factor:

n_i\propto\exp\left[-\frac{E_i}{k_BT}\right],

where $E_i=m_ic^2$ is the rest energy. Additionally, the temperature during the radiation-dominated early universe is given by $T(t)\approx10^{10}\ \mathrm K\left(\frac t{1\ \mathrm s}\right)^{-1/2}$ , where $t$ is the time since the Big Bang.

(a) (4 points) At a temperature where $k_BT\approx0.8\ \mathrm{MeV}$ , known as the freeze-out temperature, the neutrino interactions essentially stop, preventing further conversion between protons and neutrons.

i. (2 points) About how long after the Big Bang did this occur?

ii. (2 points) At the freeze-out temperature, what was the equilibrium ratio of the number density of neutrons to that of protons?

(b) (3 points) Free neutrons are unstable, and decay into protons with a characteristic decay time of $\tau=886\ \mathrm s$ (the time for which the number of neutrons drops to $1/e$ of the original amount). Given that helium nuclei only formed $t_{nuc}=200\ \mathrm s$ after freezing out, what was the ratio of the number density of neutrons to that of protons when the helium nuclei formed?

(c) (7 points) While trace amounts of several small nuclei were formed during Big Bang Nucleosynthesis (BBN), assume that all neutrons go into forming helium-4.

i. (5 points) After the helium nuclei formed, what was the ratio of the number of helium-4 nuclei to the number of hydrogen nuclei?

ii. (2 points) Approximating the mass of helium-4 as 4 times that of H (for this part only), what fraction of baryonic mass in the universe is helium?

If you weren’t able to solve part (c), assume reasonable values for the initial mass fractions of hydrogen and helium for future parts.

(d) (2 points) Albert the Astronomer claims that in older galaxies, the mass fraction of hydrogen should gradually be increasing, as neutrons slowly continue to decay into protons. Is his claim correct? If not, explain.

(e) (7 points) Suppose a certain region of a galaxy has a density of $10^{-19}\ \mathrm{kg/m^3}$ and is composed of 70% hydrogen and 30% helium-4 by mass (ignore any heavier elements). Because the region is gravitationally bound, this density doesn’t change significantly with the expansion of the universe; approximate it as constant. Assume hydrogen is converted into helium by the fusion reaction:

4\,{}^1\mathrm H^++2\mathrm e^-\to{}^4\mathrm{He}^{2+}+2\nu_e,

where the electron $\mathrm e^-$ and electron neutrino $\nu_e$ are of negligible mass. $^4\mathrm{He}$ has a mass of $m_{He}=3728.4\ \mathrm{MeV}/c^2$

i. (4 points) Over the entire time since BBN, how much energy does this process release per cubic kiloparsec? Give your answer in joules per cubic kiloparsec.

ii. (3 points) Assuming the age of the universe is 13.8 billion years, calculate the average luminosity density in solar luminosities per cubic kiloparsec.

Let’s go back and explore how we arrived at the number $t_{nuc}\approx200\ \mathrm s$ , the time at which Big Bang nucleosynthesis began. Let’s define $t_{nuc}$ as the time at which half the neutrons fused with protons into deuterium ( $^2\mathrm H$ ), as deuterium fusion is the first step in BBN. From the Maxwell-Boltzmann equation, the relative abundances of deuterium, protons and neutrons is given by

\frac{n_D}{n_pn_n}=6\left(\frac{m_nk_BT}{\pi\hbar^2}\right)^{-3/2}\exp\left(\frac{B_D}{k_BT}\right),

where $B_D=(m_p+m_n-m_D)c^2=2.22\ \mathrm{MeV}$ is the energy released in a deuterium fusion reaction.

(f) (3 points) The number density of photons is given by $n_\gamma=0.243\left(\frac{k_BT}{\hbar c}\right)^3$ . Find an expression for the number density of protons $n_p$ in terms of the temperature $T$ and the baryon to photon ratio $\eta$ . You may use your answer to part (b).

(g) (3 points) Find the present-day baryon to photon ratio. The CMB temperature is 2.725 K, and the present-day density parameter for baryonic matter is $\Omega_{b,0}=\frac{\rho_{b,0}}{\rho_{c,0}}=0.04$ . $\rho_c$ is the critical density of the universe, which is the density required for a flat universe; it is given by $\rho_c=\frac{3H^2}{8\pi G}$ . Use $H_0=70\ \mathrm{km/s/Mpc}$ .

(h) (8 points) Assuming the baryon to photon ratio is fixed since the Big Bang:

i. (5 points) Find an equation involving $T_{nuc}$ (the temperature at time $t=t_{nuc}$ ) and known constants.

ii. (1 point) What temperature $T_{nuc}$ does $t_{nuc}=200\ \mathrm s$ correspond to?

iii. (2 points) Verify that this temperature solves your equation in part (h)i.

(i) (3 points) The baryon to photon $\eta$ is a remarkably small number. One possibility is that the universe happens to prefer photons significantly over baryons. Another possibility is that a great number of quark-antiquark pairs were created in the early universe via pair production ( $\gamma+\gamma\rightleftharpoons q+\bar q$ ), and a slight asymmetry of quarks over antiquarks produced a large number of photons during quark-antiquark annihilation, leaving over a small number of quarks to form into protons and neutrons. Find the quark-antiquark asymmetry

\delta_q\equiv\frac{n_q-n_{\bar q}}{n_q+n_{\bar q}}\ll1

that would yield the baryon to photon ratio found in part (g).

Решение

(a) i. When $k_BT\approx0.8\ \mathrm{MeV}$ , we have

\begin{aligned} T&=\frac{0.8\ \mathrm{MeV}}{k_B}\\ &=\frac{1.28\cdot10^{-13}\ \mathrm J}{1.381\cdot10^{-23}\ \mathrm{J/K}}\\ &=9.28\cdot10^9\ \mathrm K. \end{aligned}

Now, we have

t=\left(\frac{10^{10}}T\right)^2=\boxed{1.16\ \mathrm s}.

ii. Let $n_p$ be the number density of protons, and $n_n$ the number density of neutrons. Then, we have that the ratio $n_n/n_p$ is

\frac{\exp[-m_nc^2/(k_BT)]}{\exp[-m_pc^2/(k_BT)]}=\exp\left[-\frac{(m_n-m_p)c^2}{k_BT}\right].

Plugging in $m_nc^2-m_pc^2=939.6-938.3=1.3\ \mathrm{MeV}$ and $k_BT=0.8\ \mathrm{MeV}$ , we get $n_n/n_p=\boxed{0.197}$ .

(b) For every proton, there is initially 0.197 of a neutron. This decays according to $n_n=n_{n,0}\exp(-t/\tau)$ . We have $0.197\exp(-200\ \mathrm s/(886\ \mathrm s))=0.157$ , so the new ratio is $\frac{0.157}{1+0.197-0.157}=\boxed{0.151}$ (taking into account that the decayed neutrons turn into protons).

(c) i. Using the answer to part (b), for every 0.151 neutrons, there is 1 proton. Then, they can form $0.151/2=0.076$ helium-4 nuclei, and the remaining $1-0.151=0.849$ protons can form 0.849 hydrogen nuclei.

Thus, the ratio of helium nuclei to hydrogen nuclei is $0.075/0.849=\boxed{0.089}$ .

ii. The mass ratio is $4\cdot0.089=0.356$ , which gives a mass percentage of $\frac{0.356}{1+0.356}=\boxed{26.3\%}$ .

(d) Albert is not correct. Only free neutrons are unstable, and the vast majority of neutrons in the universe are bound up in nuclei, particularly helium-4. Furthermore, fusion in stars actually decreases the fraction of hydrogen, as explored in the following part.

(e) i. Since the electron and electron neutrino are of negligible mass, each reaction releases $4m_pc^2-m_{He}c^2=4\cdot938.3\ \mathrm{MeV}-3728.4\ \mathrm{MeV}=24.8\ \mathrm{MeV}$ . Hydrogen went from a density of $(1-0.263)\cdot10^{-19}\ \mathrm{kg/m^3}=7.37\times10^{-20}\ \mathrm{kg/m^3}$ to $0.70\cdot10^{-19}\ \mathrm{kg/m^3}=7\times10^{-20}\ \mathrm{kg/m^3}$ , with a difference of $3.7\times10^{-21}\ \mathrm{kg/m^3}$ . This means that the energy released per cubic kiloparsec is

24.8\ \mathrm{MeV}\cdot\frac{3.7\times10^{-21}\ \mathrm{kg/m^3}}{4\cdot1.6726\times10^{-27}\ \mathrm{kg}}\cdot\left(\frac{3.086\times10^{19}\ \mathrm m}{1\ \mathrm{kpc}}\right)^3\cdot\frac{1.6022\times10^{-13}\ \mathrm J}{1\ \mathrm{MeV}}=\boxed{6.5\times10^{52}\ \mathrm{J/kpc^3}}.

ii. In seconds,13.8 billion years is $13.8\times10^9\cdot365.25\cdot24\cdot3600\ \mathrm s=4.35\times10^{17}\ \mathrm s$ . Thus, the average power per cubic kiloparsec is $\frac{6.5\times10^{52}\ \mathrm{J/kpc^3}}{4.35\times10^{17}\ \mathrm s}=1.50\times10^{35}\ \mathrm{W/kpc^3}$ . In solar luminosities, this quantity is $\frac{1.50\times10^{35}\ \mathrm{W/kpc^3}}{3.85\times10^{26}\ \mathrm{W}/L_\odot}=\boxed{3.90\times10^8\ L_\odot/\mathrm{kpc^3}}$ .

(f) From part (b), the ratio of neutrons to protons is 0.151. Thus, the proton to baryon ratio is $\frac1{1+0.151}=0.869$ , or $n_p=0.869n_b$ . By definition, $n_b=n_\gamma\eta=0.243\eta\left(\frac{k_BT}{\hbar c}\right)^3$ . We arrive at the expression

\boxed{n_p=0.211\eta\left(\frac{k_BT}{\hbar c}\right)^3}.

(g) Let us first find the present-day number density of photons. Simply plugging in $T=2.725\ \mathrm K$ into the expression given in part (f), we have

n_\gamma=0.243\left(\frac{k_B\cdot2.725\ \mathrm K}{\hbar c}\right)^3=4.09\cdot10^8\ \mathrm{photons/m^3}.

To find the present-day number density of baryons, we first need to find the present-day critical density. Using $H_0=70\ \mathrm{km/s/Mpc}=2.3\cdot10^{-18}\ \mathrm{s^{-1}}$ , we have $\rho_{c,0}=\frac{3H_0^2}{8\pi G}=9.2\cdot10^{-27}\ \mathrm{kg/m^3}$ . Since $\Omega_{b,0}=0.04$ , $\rho_{b,0}=0.04\cdot9.2\cdot10^{-27}\ \mathrm{kg/m^3}=3.7\cdot10^{-28}\ \mathrm{kg/m^3}$ . Finally, baryonic matter is composed of protons and neutrons; since $m_p\approx m_n$ , we can divide by the mass of a proton and find the number density of baryons

n_b=\frac{3.7\cdot10^{-28}\ \mathrm{kg}}{\mathrm{m^3}}\cdot\frac{1\ \mathrm{baryon}}{1.6726\cdot10^{-27}\ \mathrm{kg}}=0.22\ \mathrm{baryons/m^3}.

The baryon to photon ratio is $\eta=\boxed{5.4\cdot10^{-10}}$ .

(h) i. Earlier, we defined $t_{nuc}$ to be the time at which half the neutrons fused into deuterium, or $n_D/n_n=1$ . Setting $n_D/n_n=1$ and plugging in our expression for $n_p$ from part (f), we find

1=0.211\eta\left(\frac{k_BT}{\hbar c}\right)^3\cdot6\left(\frac{m_nk_BT}{\pi\hbar^2}\right)^{-3/2}\exp\left(\frac{B_D}{k_BT}\right)

\boxed{1\approx7\eta\left(\frac{k_BT_{nuc}}{m_nc^2}\right)^{3/2}\exp\left(\frac{B_D}{k_BT_{nuc}}\right)}

ii. Using $T(t)\approx10^{10}\ \mathrm K\left(\frac t{1\ \mathrm s}\right)^{-1/2}$ , we get $T(t=200\ \mathrm s)=\boxed{7\cdot10^8\ \mathrm K}$ . The corresponding energy is $k_BT_{nuc}=0.061\ \mathrm{MeV}$ .

iii. Using $k_BT_{nuc}=0.061\ \mathrm{MeV}$ , we have

\begin{aligned} &7\eta\left(\frac{k_BT_{nuc}}{m_nc^2}\right)^{3/2}\exp\left(\frac{B_D}{k_BT_{nuc}}\right)\\ &=7\cdot5.4\cdot10^{-10}\left(\frac{0.061\ \mathrm{MeV}}{939.6\ \mathrm{MeV}}\right)^{3/2}\exp\left(\frac{2.22\ \mathrm{MeV}}{0.061\ \mathrm{MeV}}\right)=13. \end{aligned}

Due to the exponential, this expression is very sensitive to small changes in $k_BT$ . Thus for this expression, an answer of 13 is roughly consistent with 1. The exact solution is $k_BT=0.066\ \mathrm{MeV}$ , which still corresponds to $t_{nuc}=200\ \mathrm s$ to the nearest significant figure.

(i) $2n_{\bar q}$ quarks are annihilated, producing $2n_{\bar q}\approx n_q+n_{\bar q}$ photons. $n_q-n_{\bar q}$ quarks are left over to form $(n_q-n_{\bar q})/3$ baryons. The resulting baryon to photon ratio is thus

\eta=\frac{(n_q-n_{\bar q})/3}{n_q+n_{\bar q}}=\frac13\delta_q.

Using $\eta=5.4\cdot10^{-10}$ , we find a quark-antiquark asymmetry of $\boxed{\delta_q=1.6\cdot10^{-9}}$ ; there was one extra quark in 800 million quark-antiquark pairs.

Source note: The original key (c) first gives 0.076 helium nuclei but uses 0.075 in the following numerator; both printed values are preserved. In (h)iii the key accepts 13 as roughly consistent with 1, explains the exponential sensitivity, and gives 0.066 MeV as the exact solution. These original numerical claims and marking explanation are retained.

Теми и трудност

Оценена трудност: 5/5.

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Оригинал в Архива: usaaao_second_exam_2021_final.pdf · официални решения: usaaao_second_exam_2021_solutions.pdf · допълнителни данни: physical-constants.pdf