IOAA 2015, Теория, кратки задачи, английски — Задача 12

SHORT PROBLEMS

Условие

A proton with a kinetic energy of 1GeV1GeV propagates out from the surface of the Sun towards the Earth. Neglecting the magnetic field of the Sun, calculate the travel time of the proton as seen from the Earth.

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