IOAA 2015, Теория, кратки задачи, английски — Задача 6

SHORT PROBLEMS

Условие

At the start of every observation, a radio telescope is pointed at a point-source calibrator that has a known flux density of 21.86 Jy outside the Earth’s atmosphere. However, on a certain date, the measured flux density of the calibrator source was 14.27 Jy. If the calibrator source was at an altitude of 35 degrees, estimate the zenith atmospheric optical depth, τz\tau_z.

Решение

13. Earlier marking-scheme variant
  1. 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+ΔE→D+e++νep + p + \Delta E \rightarrow D + e^{+} + \nu_e [20]

2 ΔE=(me+mD−2mp)c2=(me+(mp+mn)−2mp)c2=(me+mn−mp)c2\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 ΔE=(9.109×10−31+1.675×10−27−1.673×10−27)kg (2.99792×108ms−1)2=2.6198×10−13 J\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 ΔE∼kBT→T∼ΔEkB\Delta E \sim k_BT \rightarrow T \sim \frac{\Delta E}{k_B} [20]

5 T=2.6198×10−13 J8.315Jmol K=2.6198×10−13mol K8.315×6.02×10231mol=1.894×1010 KT = \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