Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 59

Verify that the total number of nucleons, and total charge are conserved for each of the following fusion reactions in the proton-proton chain. (i) \(^{1} \mathrm{H}+^{1} \mathrm{H} \rightarrow^{2} \mathrm{H}+e^{+}+v_{\mathrm{e}},\) (ii) \(^{1} \mathrm{H}+^{2} \mathrm{H} \rightarrow^{3} \mathrm{He}+\gamma,\) and (iii) \(^{3} \mathrm{He}+^{3} \mathrm{He} \rightarrow^{4} \mathrm{He}+^{1} \mathrm{H}+^{1} \mathrm{H}\). (List the value of each of the conserved quantities before and after each of the reactions.)

Short Answer

Expert verified
In all three fusion reactions in the proton-proton chain, the total number of nucleons and total charge are conserved: (i) \(^{1}H + ^{1}H \rightarrow ^{2}H + e^{+} +v_{e}\): nucleons (2) and charge (+2) conserved. (ii) \(^{1}H + ^{2}H \rightarrow ^{3}He + \gamma\): nucleons (3) and charge (+2) conserved. (iii) \(^{3}He + ^{3}He \rightarrow ^{4}He + ^{1}H + ^{1}H\): nucleons (6) and charge (+4) conserved.

Step by step solution

01

(i) First Reaction

\(^{1}H + ^{1}H \rightarrow ^{2}H + e^{+} +v_{e}\) Before the reaction: Total number of nucleons = Number of protons + Number of neutrons = 1 + 1 = 2 Total charge = Charge of Protons = +1 +1 = +2 After the reaction: Total number of nucleons = Number of protons + Number of neutrons = 1 + 1 = 2 Total charge = Charge of Deuteron + Charge of positron = +1 + 1 = +2 From the before and after calculations, we can see that the total number of nucleons (2) and total charge (+2) are conserved.
02

(ii) Second Reaction

\(^{1}H + ^{2}H \rightarrow ^{3}He + \gamma\) Before the reaction: Total number of nucleons = Number of protons + Number of neutrons = 1 + 2 = 3 Total charge = Charge of Protons = +1 +1 = +2 After the reaction: Total number of nucleons = Number of protons + Number of neutrons = 2 + 1 = 3 Total charge = Charge of Helium-3 = +2 From the before and after calculations, we can see that the total number of nucleons (3) and total charge (+2) are conserved.
03

(iii) Third Reaction

\(^{3}He + ^{3}He \rightarrow ^{4}He + ^{1}H + ^{1}H\) Before the reaction: Total number of nucleons = Number of protons + Number of neutrons = 3 + 3 = 6 Total charge = Charge of Protons = +2 + 2 = +4 After the reaction: Total number of nucleons = Number of protons + Number of neutrons = 2 + 2 + 1 + 1 = 6 Total charge = Charge of Helium-4 + Charge of Protons = +2 +1 +1 = +4 From the before and after calculations, we can see that the total number of nucleons (6) and total charge (+4) are conserved.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Explain why a bound system should have less mass than its components. Why is this not observed traditionally, say, for a building made of bricks?

Silver has two stable isotopes. The nucleus, \(^{107}_{47} \mathrm{Ag},\) has atomic mass \(106.905095 \mathrm{g} / \mathrm{mol}\) with an abundance of \(51.83 \% ;\) whereas \(^{109}_{47} \mathrm{Ag}\) has atomic mass \(108.904754 \mathrm{g} /\)mol with an abundance of \(48.17 \% .\) Find the atomic mass of the element silver.

Calculate the activity \(R,\) in curies of \(1.00 \mathrm{g}\) of \(^{226} \mathrm{Ra}\). (b) Explain why your answer is not exactly 1.00 Ci, given that the curie was originally supposed to be exactly the activity of a gram of radium.

A large power reactor that has been in operation for some months is turned off, but residual activity in the core still produces \(150 \mathrm{MW}\) of power. If the average energy per decay of the fission products is \(1.00 \mathrm{MeV}\), what is the core activity?

Another set of reactions that fuses hydrogen into helium in the Sun and especially in hotter stars is called the CNO cycle: $$\begin{aligned} &^{12} \mathrm{C}+^{1} \mathrm{H} \rightarrow^{13} \mathrm{N}+\gamma\\\ &^{13} \mathrm{N} \rightarrow^{13} \mathrm{C}+e^{+}+v_{\mathrm{e}}\\\ &^{13} \mathrm{C}+^{1} \mathrm{H} \rightarrow^{14} \mathrm{N}+\gamma\\\ &^{14} \mathrm{N}+^{1} \mathrm{H} \rightarrow^{15} \mathrm{O}+\gamma\\\ &^{15} \mathrm{O} \rightarrow^{15} \mathrm{N}+e^{+}+v_{\mathrm{e}}\\\ &^{15} \mathrm{N}+^{1} \mathrm{H} \rightarrow^{12} \mathrm{C}+^{4} \mathrm{He} \end{aligned}$$ This process is a "cycle" because \(^{12} \mathrm{C}\) appears at the beginning and end of these reactions. Write down the overall effect of this cycle (as done for the proton-proton chain in \(2 e^{-}+4^{1} \mathrm{H} \rightarrow^{4} \mathrm{He}+2 v_{\mathrm{e}}+6 \gamma\) ). Assume that the positrons annihilate electrons to form more \(\gamma\) rays.
See all solutions

Recommended explanations on Physics Textbooks

Classical Mechanics

Read Explanation

Fundamentals of Physics

Read Explanation

Dynamics

Read Explanation

Quantum Physics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Medical Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free