Съдържание

IOAA 2015, Теория, кратки задачи, английски — Задача 4

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от IOAA 2015

SHORT PROBLEMS

Условие

Let us assume that an observer using a hypothetical, far-infrared, Earth-sized telescope (wavelength range 20 to 640 μm) found a static and neutral supermassive black hole with a mass of 2.1 x $10^{10}$ $\mathrm{M_\odot}$ . Determine the maximum distance at which this black hole can be resolved by the observer.

Решение

Answer and Marking Scheme:

1 Identify the relationship between the telescope’s resolution and the angular size of black hole

[25]

This problem can be solved by using relationship between the telescope’s resolution and the angular size of black holes which can be seen in this equation:

\phi_{telescope} \le \phi_{blackhole} \;\Rightarrow\; 1.2\frac{\lambda}{D_{telescope}} \le \frac{\text{Diameter of event horizon}}{\text{Observer Distance to galatic' s the center}}

1.2\frac{\lambda}{D_{telescope}} \le \frac{2R_{blackhole}}{d}

2 Identify the relation for determining the distance

[50]

Then, it is assumed that the radius of black hole is the Schwarszchild radius; and the telescope is also Earth-sized telescope. Therefore:

1.2\frac{\lambda}{2R_{Earth}} \le \frac{2R_{Schwarzchild,BH}}{d}

d \le \frac{4\times R_{Schwarzchild,BH}\times R_{Earth}}{1.2\times\lambda} \;\Rightarrow\; d \le \frac{4\times M_{Black\ Hole}\times R_{Schwarchild,Sun}\times R_{Earth}}{1.2\times M_{Sun}\times\lambda}

d \le \frac{4\times2.1\times10^{10}M_{Sun}\times R_{Schwarchild,Sun}\times R_{Earth}}{1.2\times M_{Sun}\times\lambda} \;\Rightarrow\; d \le \frac{8.4\times10^{10}\times R_{Schwarchild,Sun}\times R_{Earth}}{1.2\times\lambda}

3 Calculating the maximum-observer distance

[25]

To find the maximum distance, then we should use the minimum λ.

Therefore:

d_{\max} = \frac{8.4\times10^{10}\times R_{Schwarchild,Sun}\times R_{Earth}}{1.2\times\lambda_{\min}} = \frac{8.4\times10^{10}\times2.95\times10^{3}\times6.3708\times10^{6}}{1.2\times20\times10^{-6}}\ m = 6.6\times10^{25}\ m

13. Earlier marking-scheme variant

Two protons collide to produce deuteron, positron, and neutrino in the center of a star. Write down the reaction equation and calculate the minimum temperature needed for this reaction. Assume that the neutrino mass is negligible.

Answer

1 $p + p + \Delta E \rightarrow D + e^{+} + \nu_e$ [20]

2 $\Delta E = (m_e + m_D - 2m_p)c^2 = (m_e + (m_p + m_n) - 2m_p)c^2 = (m_e + m_n - m_p)c^2$ [20]

3 $\Delta E = (9.109\times10^{-31} + 1.675\times10^{-27} - 1.673\times10^{-27})\mathrm{kg}\,(2.99792\times10^{8}\mathrm{ms^{-1}})^2 = 2.6198\times10^{-13}\ J$ [20]

4 $\Delta E \sim k_BT \rightarrow T \sim \frac{\Delta E}{k_B}$ [20]

5 $T = \frac{2.6198\times10^{-13}\ J}{8.315\frac{J}{mol\ K}} = \frac{2.6198\times10^{-13}mol\ K}{8.315}\times6.02\times10^{23}\frac{1}{mol} = 1.894\times10^{10}\ K$ [20]

Оригинал в Архива: Short_problem_Final_EN.pdf · официални решения: Short Problem Solution-Marking Scheme 30.07.pdf · допълнителни данни: Short_Problem_Solution.docx