Home Page Textbooks Physics: Principles and Problems Page 814: Section Review

Physics: Principles and Problems

9th Edition

Elliott, Haase, Harper, Herzog, Margaret Zorn, Nelson, Schuler, Zitzewitz

ISBN: 9780078458132

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Page 814: Section Review

Exercise 28

Step 1

1 of 1

The reason for this, at first sight, paradoxical behavior is the fact that neutrons are heavier than protons and during the transmutation, this difference in mass (energy) is used to produce an electron and an antineutrino.

Exercise 29

Step 1

1 of 4

In this problem, we are asked to write the nuclear equation for $alpha$-decay of the polonium isotope $^{210}_{84}$Po.

Step 2

2 of 4

In order to do so, we are going to write the general $alpha$-decay equation which is given as

$$
^{A}_ZXrightarrow^{A-4}_{Z-2}Y+^4_2alpha
$$

Step 3

3 of 4

From the equation above we see that $Z-2=82$ so the child nucleus is lead. We can write then

$$
^{210}_{84}textrm{Po}rightarrow^{206}_{82}textrm{Pb}+^4_2alpha
$$

Result

4 of 4

$$
^{210}_{84}textrm{Po}rightarrow^{206}_{82}textrm{Pb}+^4_2alpha
$$

Exercise 30

Step 1

1 of 2

From Figure 30-4 we see that the activity drops down to 3/8 of the original value after $approx1.5$ half-lives. From the table 30-2 we see that the $^{131}_{53}$I half-life is 8.07 days so we can expect that this activity would be reached after $boxed{approx12}$ days.

Result

2 of 2

12 days

Exercise 31

Step 1

1 of 1

Lead wouldn’t be a good nuclear moderator since it would absorb neutrons non-selectively. Controlled chain reactions work in such way that we need a neutrons to have energies up to a threshold which allows for the control. If they are above that, they get absorbed by the moderator material.

Exercise 32

Step 1

1 of 1

The reason for this is that inside the molecule, deuterium nuclei can be considered static relative to each other. Therefore, they do field an electrostatic repulsion since they both have a positive net charge. To overcome this strong repulsion nuclei should have very high speeds, i.e. high kinetic energies which would lead to the fusion events upon collisions.

Exercise 33

Step 1

1 of 4

In this problem we are given the first link in the chain of fusion reactions in the Sun. We should find the energy gain from this reaction.

Step 2

2 of 4

In order to find this energy, we should know what is the mass defect of the reaction and then multiply it by the conversion factor. Since the mass of the neutrino can be neglected, we have that

$$
m_d=2times m_H-m_D-m_{e^+}=2times 1.007825-2.014102-0.00054858
$$

$$
m_d=0.00099942textrm{ u}
$$

Step 3

3 of 4

center{Now, the energy gain can be obtained as }
[E=931.5times m_d=931.5times 0.00099942=boxed{0.931textrm{ MeV}}]

Result

4 of 4

$$
E=0.931textrm{ MeV}
$$

Exercise 34

Step 1

1 of 1

By definition, more positive potential point will have more positively charged particles than the reference point. In this case, the plate that is struck by alpha particles will be at higher potential since it will have more positively charged particles than the other plate.

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