What is the average kinetic energy of protons at the center of a star where the temperature is \(1.00 \cdot 10^{7} \mathrm{~K} ?\) What is the average velocity of those protons?

Short Answer

Expert verified
Answer: The average kinetic energy of protons is \(2.07 \cdot 10^{-16} \mathrm{~J}\), and their average velocity is \(4.98 \cdot 10^5 \mathrm{~m/s}\).

Step by step solution

01

Calculate the average kinetic energy of protons

Using the Equipartition theorem, the average kinetic energy per particle is given by: \(KE = \frac{3}{2}kT\) Given temperature \(T = 1.00 \cdot 10^{7}\mathrm{~K}\) and Boltzmann constant \(k = 1.38 \cdot 10^{-23} \mathrm{~J / K}\), we can find the average kinetic energy as: \(KE = \frac{3}{2} \times 1.38 \cdot 10^{-23} \mathrm{~J / K} \times 1.00 \cdot 10^{7} \mathrm{~K} = 2.07 \cdot 10^{-16} \mathrm{~J}\)
02

Calculate the average velocity of protons

Now that we have the average kinetic energy, we can use the kinetic energy formula \(\frac{1}{2}mv^2\) to find the average velocity of protons: \(2.07 \cdot 10^{-16} \mathrm{~J} = \frac{1}{2}(1.67 \cdot 10^{-27} \mathrm{~kg})(v^2)\) Solving for \(v\), we get: \(v^2 = \frac{2(2.07 \cdot 10^{-16} \mathrm{~J})}{1.67 \cdot 10^{-27} \mathrm{~kg}}\) \(v^2 = 2.48 \cdot 10^{11} \mathrm{~m^2/s^2}\) \(v = \sqrt{2.48 \cdot 10^{11} \mathrm{~m^2/s^2}}\) \(v = 4.98 \cdot 10^5 \mathrm{~m/s}\) So, the average velocity of protons at the center of the star is \(4.98 \cdot 10^5 \mathrm{~m/s}\).

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