Съдържание

USAAAO 2018, II кръг — Задача 14

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от USAAAO 2018

2018 National Astronomy Olympiad (NAO) · 15 т.

Instructions (Please Read Carefully)

(Start the exam after reading the instructions carefully and affirming that the contestant understood the instructions given.)

The top 5 eligible scorers on the NAO 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 2 parts: Short Questions (5 points each) and Long Questions (15 points each). The maximum score is 100 points.

The test must be completed within 2.5 hours (150 minutes). The proctor should mark the start and end time of the exam on the front page.

Please solve each problem on a blank piece of paper and mark the number of the problem at the top of the page. The contestants 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 contestants 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 exam document, solution, and all the scratch paper used should be turned in to the proctor at the end of the exam.

This is a written exam. Contestants can only use a scientific 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 6th. Receiving any external help during the exam is strictly prohibited. This means that the only allowed items are: a scientific calculator, the provided table of constants, a pencil (or pen), an eraser, blank sheets of papers, and the exam. No books, notes, laptops, mobile phones, or any other devices are allowed while taking the exam.

After reading the instructions, please make sure to sign at the bottom of this page, affirming that:

All work on this exam is mine.
I took this exam under a proctors supervision.
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.

Условие

Suppose that at some time in the very recent past all the hydrogen and helium (baryon density $\rho_b = 4.2 \times 10^{-31}$ g/cm $^{3}$ , about 75% hydrogen (=1 baryon) by mass and 25% helium (=4 baryons) by mass) in the universe had been instantly fused into iron in stars, and the released energy thermalized into black body radiation. (Note that the binding energy per nucleon of ${}^{56}_{26}$ Fe is 8.8 MeV and that of ${}^{4}_{2}$ He is 7.1 MeV.)

(a) Calculate the current temperature of this black body radiation. [5 т.]

(b) At what wavelength would the black body spectrum peak, and what region of the electromagnetic spectrum would this be? [5 т.]

(c) The mean bolometric luminosity per unit volume emitted by stars in the universe today is about $3 \times 10^{8}\ L_{\odot}/\mathrm{Mpc}^3$ . How long would it takes stars at this rate to fuse all the hydrogen and helium in the universe? Compare to the present age of the universe. [5 т.]

Отговори

Решение

(a) Answer: $T = 4.40K$ .

Without considering intermediary products, we have these two reactions:

56\ {}^{1}_{1}\mathrm{H} \rightarrow {}^{56}_{26}\mathrm{Fe}

14\ {}^{4}_{2}\mathrm{He} \rightarrow {}^{56}_{26}\mathrm{Fe}

We can calculate the energy released in each of these reactions by computing the binding energy on each side. There is no binding energy for a hydrogen atom, so the energy released in the first reaction is $56 \times 8.8$ MeV = 492.8 MeV per Fe. For the second reaction, the binding energy will by ( $56 \times 8.8$ MeV) - ( $14 \times 4 \times 7.1$ MeV) = 95.2 MeV per Fe.

To find how much iron is actually produced, we need to determine the number densities of hydrogen and helium using the mass fraction given in the question:

n_H = \frac{0.75\rho_b}{m_H}

n_{He} = \frac{0.25\rho_b}{m_{He}}

where $m_H$ and $m_{He}$ are the masses of hydrogen and helium respectively. Then , the total energy density released is:

u = \frac{492.8MeV \times n_H}{56} + \frac{95.2MeV \times n_{He}}{14}

To get the temperature, we use:

u = aT^4

where $a$ is the radiation constant. This gives us a temperature of 4.40 K.

(b) Answer: $\lambda = 6.59 \times 10^{-2}$ cm, microwave.

Using Wien’s displacement law:

\lambda = \frac{b}{T}

where $b = 2.898 \times 10^{-3}$ m K, we get a wavelength of $6.59 \times 10^{-2}$ cm, which is in the microwave region of the electromagnetic spectrum.

(c) Answer: $7.11 \times 10^{19}$ seconds (2250 Gyr)

The time it would take is given by the energy density (found in part (a)), divided by the rate of fusion, i.e. the luminosity given in the problem:

t = \frac{u}{L}

This gives us $7.11 \times 10^{19}$ seconds, or about 2250 Gyr. This is much longer than the current age of the universe (13.7 Gyr).

Оригинал в Архива: usaaao-nao-2018.pdf · официални решения: usaaao_second_round_2018.pdf