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Chapter 38: Problem 65

The proton-proton cycle consumes four protons while producing about 27 MeV of energy. (a) At what rate must the Sun consume protons to produce its power output of about \(4 \times 10^{26} \mathrm{W} ?\) (b) The present phase of the Sun's life will end when it has consumed about \(10 \%\) of its original protons. Estimate how long this phase will last, assuming the Sun's \(2 \times 10^{30}-\mathrm{kg}\) mass was initially \(71 \%\) hydrogen.

Short Answer

Expert verified
The rate of proton consumption is \(4 \times 10^{26} J/s / 1.08 \times 10^{-12} J/proton \approx 3.7 \times 10^{38} protons/s\). The total number of protons is \(0.71 \times 2 \times 10^{30} kg / 1.67 \times 10^{-27} kg/proton \approx 8.5 \times 10^{56} protons\). The Sun will last for \(0.10 \times 8.5 \times 10^{56} protons / 3.7 \times 10^{38} protons/s \approx 2.3 \times 10^{17} s \), which is approximately 7.3 billion years in term of the sun's lifetime.

Step by step solution

01

Calculate rate of proton consumption

To find the rate at which the Sun consumes protons, the provided total energy output of the Sun, which is \(4 \times 10^{26} \mathrm{W}\), should be divided by the energy output per proton. As described in the exercise the energy produced by consuming 4 protons, is about 27MeV, or \(27 \times 10^6\) electronvolt. 1MeV equals \(1.6 \times 10^{-19}\) joules, so the energy per reaction in joules is \(27 \times 10^6 \times 1.6 \times 10^{-13} \mathrm{J}\). Since 4 protons get consumed in one reaction, the energy per proton is \(27 \times 10^6 \times 1.6 \times 10^{-19} / 4 \mathrm{J}\). Dividing the total energy output of the Sun by the energy per proton will yield the rate of proton consumption per second.
02

Calculate total number of protons

To calculate the time before the Sun has consumed 10% of its original protons, it's essential to find out the total initial quantity of protons. As given, the Sun's original mass was \(2 \times 10^{30} kg\), out of which 71% was hydrogen. This means the initial hydrogen mass is \(0.71 \times 2 \times 10^{30} kg\). Since hydrogen's mass is approximately equal to the mass of a single proton, conversion to the number of protons can be made using the mass of a proton which is approximately \(1.67 x 10^{-27} kg\). Hence, the number of protons initially was \(0.71 \times 2 \times 10^{30} / 1.67 \times 10^{-27}\).
03

Determine how long the Sun will last

To find the time for which the sun can continue to burn before it has consumed 10% of its original protons, the number of protons initially (as calculated in previous step) should be multiplied by 0.10. Then, this value is divided by the rate of consumption of protons per second (as calculated in the first step). The result will be in seconds, so further conversion could be desired depending on the wanted units.

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Most popular questions from this chapter

Passage Problems In \(1972,\) a worker at a nuclear fuel plant in France discovered that uranium from a mine in Oklo, in the African Republic of Gabon, had less U-235 than the normal 0.7 \(\%\) - a quantity known from meteorites and Moon rocks to be constant throughout the solar system. Further analysis showed the presence of isotopes that would result from the decay of fission products. Scientists drew the remarkable conclusion that a natural nuclear fission reaction had occurred some 2 billion years ago, lasting for about 100,000 years. Water, mixing with rich uranium ore, provided the moderator that made the chain reaction possible. More significantly, U-235's 700-million-year half-life means that 2 billion years ago there was a higher abundance of \(U-235\) in natural uranium. At the time of the Oklo fission reaction, the actual amount of U-235 present was a. about the same as today. b. about twice as much as today. c. about four times as much as today. d. about eight times as much as today.

(a) Example 38.6 explains that the number of fission events in a chain reaction increases by a factor \(k\) with each generation. Show that the total number of fission events in \(n\) generations is \(N=\left(k^{n+1}-1\right) /(k-1) \cdot(b)\) In a typical nuclear explosive, \(k\) is about 1.5 and the generation time is about \(10 \mathrm{ns}\). Use the result from (a) to calculate the time for all the nuclei in a 10 -kg mass \(^{235} \mathrm{U}\) to fission. (Hint: Sum a series in part (a), and neglect 1 compared with \(N\) in part (b).)

Show that the decay constant and half-life are related by \(t_{1 / 2}=\ln 2 / \lambda \simeq 0.693 / \lambda.\)

A family member is about to have a brain scan using technetium\(99^{*},\) an excited isotope with 6.01 -hour half-life. The hospital makes Tc-99* from the decay of molybdenum-99 \(\left(t_{1 / 2}=2.7 \text { days }\right)\) then delivers it to the nuclear medicine department. You're told that the \(\mathrm{Tc}-99^{*}\) will arrive 90 minutes after production, and that there must be 10 mg of it. The technician says she will produce \(12 \mathrm{mg}\) of \(\mathrm{Tc}-99^{* *} .\) Is that sufficient?

On an energy-release-per-unit-mass basis, by approximately what factor do nuclear reactions exceed chemical reactions?
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