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Chapter 40: Problem 51

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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Most popular questions from this chapter

Consider a 42.58 -MHz photon needed to produce NMR transition in free protons in a magnetic field of \(1.000 \mathrm{~T}\). What is the wavelength of the photon, its energy, and the region of the spectrum in which it lies? Could it be harmful to the human body?

The most common isotope of uranium, \({ }_{92}^{238} \mathrm{U},\) produces radon \({ }_{86}^{222} \mathrm{Rn}\) through the following sequence of decays: $$\begin{array}{c}{ }^{238} \mathrm{U} \rightarrow{ }^{234} \mathrm{Th}+\alpha,{ }^{234} \mathrm{Th} \rightarrow{ }^{234} \mathrm{~Pa}+\beta^{-}+\bar{\nu}_{e}, \\\\{ }_{91}^{234} \mathrm{~Pa} \rightarrow{ }_{92}^{234} \mathrm{U}+\beta+\bar{\nu}_{e},{ }^{234} \mathrm{U} \rightarrow{ }^{230} \mathrm{Th}+\alpha ,\\\\{ }_{91}^{230} \mathrm{Th} \rightarrow{ }_{90}^{226} \mathrm{Ra}+\alpha,{ }_{88}^{226} \mathrm{Ra} \rightarrow{ }_{86}^{222} \mathrm{Rn}+\alpha,\end{array}$$. A sample of \({ }_{92}^{238} \mathrm{U}\) will build up equilibrium concentrations of its daughter nuclei down to \({ }_{88}^{226} \mathrm{Ra} ;\) the concentrations of each are such that each daughter is produced as fast as it decays. The \({ }_{88}^{226} \mathrm{Ra}\) decays to \({ }_{86}^{222} \mathrm{Rn},\) which escapes as a gas. (The \(\alpha\) particles also escape, as helium; this is a source of much of the helium found on Earth.) In high concentrations, the radon is a health hazard in buildings built on soil or foundations containing uranium ores, as it can be inhaled. a) Look up the necessary data, and calculate the rate at which \(1.00 \mathrm{~kg}\) of an equilibrium mixture of \({ }_{92}^{238} \mathrm{U}\) and its first five daughters produces \({ }_{86}^{222} \mathrm{Rn}\) (mass per unit time). b) What activity (in curies per unit time) of radon does this represent?

The binding energy of \({ }_{2}^{3}\) He is lower than that of \({ }_{1}^{3} \mathrm{H} .\) Provide a plausible explanation, considering the Coulomb interaction between two protons in \({ }_{2}^{3}\) He.

The strong force (select all that apply.) a) is only attractive. b) does not act on electrons. c) only acts over a few \(\mathrm{fm}\). d) All of the above are true. e) None of the above are true.

Before you look it up, make a prediction of the spin (intrinsic spin, i.e., actual angular momentum) of the deuteron, \({ }^{2} \mathrm{H}\). Explain your reasoning.
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