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2023 National Astronomy Competition 1 Instructions (Please Read Carefully)Table of Physical Constants Astrophysical Data Other data and conversion factors Trigonometrical identities Prefixes Условие Решение

USAAAO 2023, II кръг — Задача 4. Medium 1 · Hyperfine splitting

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2023 National Astronomy Competition · 20 т.

Внимание: Бележка към темата

The separately attributed physical-constants.pdf is an undated common USAAAO archive reference; its identity as the table supplied for this exam is unproven. Printed question/key inconsistencies are preserved, including Medium 3 P5 radius and graph slope, Long 1 cosine denominator, and Long 2 exit-coordinate term and proper-motion magnitude.

Instructions (Please Read Carefully)

2023 National Astronomy Competition

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 240 points.

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 April 1st. 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.

We acknowledge the following people for their contributions to this year’s exam:

Wesley Antônio Machado Andrade de Aguiar, Erez Abrams, Lucas Pinheiro, Sahil Pontula, Joe McCarty, Hagan Hensley, Leo Yao, and Andrew Liu.

Undated common archive reference · Table of Physical Constants

Table of Physical Constants

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 ( $\vphantom{C}^{12}\mathrm C$ 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

Undated common archive reference · Astrophysical Data and Conversion Factors

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	$1\,\mathrm{N\,m^{-2}}$
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

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

Условие

(20 points) Hyperfine splitting

The 21cm spectral line of hydrogen is a result of the interaction between the electron’s and proton’s quantum mechanical spin (known as hyperfine splitting). The spins can be either aligned or anti-aligned. When a hydrogen atom decays from the higher energy state to the lower energy state, a photon is emitted with energy equal to the energy difference in these two states.

The intrinsic magnetic moment of the electron $\mu_e$ is approximately equal to the Bohr magneton $\mu_B=\frac{e\hbar}{2m_e}$ , and the intrinsic magnetic moment of the proton $\mu_p$ is roughly $2.8\frac{e\hbar}{2m_p}$ . Electrons and protons (and any particles with spin $\frac12$ ) have permanent magnetic dipole moments $\vec m$ with magnitude equal to the intrinsic magnetic moment.

(a) (5 points) Consider two classical magnetic dipoles $\vec m_1,\vec m_2$ separated by a distance $\vec r=r\hat r$ . The magnetic field from a perfect magnetic dipole $\vec m$ is

\vec B=\frac{\mu_0}{4\pi}\frac{3\hat r(\vec m\cdot\hat r)-\vec m}{r^3}

The potential energy of a magnetic dipole in an external magnetic field $\vec B$ (e.g. the field created by the other dipole) is $U=-\vec m\cdot\vec B$ .

Using these two formulas together, what is the energy $U_{\mathrm{int}}$ of the interaction between two magnetic dipoles $\vec m_1,\vec m_2$ separated by a distance $\vec r=r\hat r$ ?

(b) (6 points) In our case, we can assume that $r$ is the Bohr radius $a_0$ , and both $\vec m_1$ and $\vec m_2$ are oriented perpendicularly to $\vec r$ . What is the interaction energy $U_{\mathrm{int}}$ between the magnetic moments of the proton and the electron? Write your answer in terms of $\vec m_p\cdot\vec m_e$ .

(c) (2 points) Which state is the lower energy state: the state where the spins are aligned, or the state where the spins are anti-aligned?

(d) (7 points) The energy of the photon emitted by the transition between these two states is equal to the energy difference in the two states. What value would you predict for the wavelength of the 21cm spectral line? (You should get the right answer to within an order of magnitude, but note that we are treating a quantum mechanical system classically and so there will be large errors.)

Решение

Solution

(a) The interaction energy between the pair of magnetic dipoles is equal to the potential energy of one of the dipoles due to the other dipole’s magnetic field. We can write this as

U_{\mathrm{int}}=-\vec m_1\cdot\vec B(\vec m_2)=-\vec m_1\cdot\frac{\mu_0}{4\pi}\frac{3\hat r(\vec m_2\cdot\hat r)-\vec m_2}{r^3}

=-\frac{\mu_0}{4\pi}\frac{3(\vec m_1\cdot\hat r)(\vec m_2\cdot\hat r)-\vec m_1\cdot\vec m_2}{r^3}

It’s good to check - this expression is symmetrical in $\vec m_1$ and $\vec m_2$ , so it didn’t matter which of the dipoles we used to do the calculation.

(b) Since $\vec m_1$ and $\vec m_2$ are perpendicular to $\vec r$ , the $\vec m\cdot\vec r$ terms vanish and the expression simplifies dramatically. If we plug in $a_0$ for the radius $r$ , then the interaction energy becomes:

U_{\mathrm{int}}=\frac{\mu_0}{4\pi a_0^3}\vec m_p\cdot\vec m_e

(c) $\vec m_p\cdot\vec m_e$ is positive if the spins are aligned and negative if they are anti-aligned. Since $U_{\mathrm{int}}$ is proportional to $\vec m_p\cdot\vec m_e$ , we can see that the higher energy state has the spins aligned while the lower energy state has the spins anti-aligned.

(d) The magnitude of $\vec m$ for the electron or proton is just equal to the intrinsic magnetic moment. This means that the aligned state has interaction energy

E_+=\frac{\mu_0}{4\pi a_0^3}\mu_e\mu_p

and the anti-aligned state has interaction energy

E_-=-\frac{\mu_0}{4\pi a_0^3}\mu_e\mu_p

The energy of the emitted photon is equal to the energy difference in these two states, which is

\Delta E=E_+-E_-=2\frac{\mu_0}{4\pi a_0^3}\mu_e\mu_p=\frac{\mu_0}{4\pi a_0^3}\frac{2.8(e\hbar)^2}{2m_em_p}=1.1\cdot10^{-6}\ \mathrm{eV}

The energy of a photon is $E=\frac{hc}\lambda$ , so the corresponding wavelength of this photon is 112cm.

(As it turns out, this is off from the correct value by a factor of exactly $\frac{16}3$ , which can only be explained by quantum mechanics. Specifically, the wavefunctions of the nucleus and electron actually overlap, whereas we assumed they were discrete point particles with a well-defined separation.)

Оригинал в Архива: second_exam_2023-1.pdf · официални решения: second_exam_2023_sols-1.pdf · допълнителни данни: physical-constants.pdf