IOAA 2015, Теория, кратки задачи, английски — Задача 11

SHORT PROBLEMS

Условие

The mass density of an object is inversely proportional to the radial distance from the center of the object with a factor of proportionality α=5.0×1013 kg/m2\alpha = 5.0\times10^{13}\ kg/m^2. If the escape velocity at the surface of the object is v0=1.5×104 m/sv_0 = 1.5\times10^{4}\ m/s , calculate the total mass of the object.

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