Physics: Principles and Problems
9th Edition
ISBN: 9780078458132
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Page 805: Section Review
Exercise 9
Step 1
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In this problem we are given two pairs of nuclei: 1) $^{12}_6$C and $^{13}_6$C one one side and 2) $^{11}_5$B and $^{11}_6$C. We want to discus their differences and their similarities.
Step 2
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The first pair has the same atomic number, i.e. it is essentially the same element, carbon, but it has different number of nucleons, i.e. neutrons.
Step 3
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The second pair nuclei have different atomic numbers but the same number of nucleons.
Exercise 10
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By all means, it is the tritium that has higher binding energy because it will have a larger mass defect. The reason for this is that tritium has 2 neutrons and $^3_2$He has only one and neutrons are heavier than protons.
Result
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Tritium has the more negative binding energy.
Exercise 11
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The reason for this is that at certain cutoff distance, the strong nuclear force will decline sharply whereas the repulsion that protons feel from each other will remain and at certain point it can prevail. It is important to know that electrostatic repulsion is inversely proportional to the square of distance and at very small distance can be significant.
Exercise 12
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Since the tritium has a neutron more, it has a larger mass defect since neutrons are heavier than protons.
Exercise 13
Step 1
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In this problem, we are given a radioactive carbon isotope, $^{14}_6$C and its mass. We want to know the mass defect and the corresponding binding energy of the isotope.
Step 2
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The mass defect is defined as the difference between the mass of the atom and the mass of it’s constituents and it can be written as follows
$$
M_d=M-Ztimes(m_p+m_e)-(A-Z)times m_n
$$
Step 3
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center{a) In our case we have that }
[M_d=M-6(m_p+m_e)-8m_n=14.003074-6times 1.007825-8times 1.008665]
[boxed{M_d=-0.113196textrm{ u}}]
[M_d=M-6(m_p+m_e)-8m_n=14.003074-6times 1.007825-8times 1.008665]
[boxed{M_d=-0.113196textrm{ u}}]
Step 4
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b) The binding energy in MeV is obtained when the mass defect is multiplied by the conversion factor ($1u=931.5$MeV) so that we have
$$
E_B=931.5times M_d=-931.5times 0.113196=boxed{-105.44textrm{ MeV}}
$$
Result
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$$
textrm{a) }M_d=-0.113196textrm{ u}
$$
textrm{a) }M_d=-0.113196textrm{ u}
$$
$$
textrm{b) }E_B=-105.44textrm{ MeV}
$$
Exercise 14
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If we observe the Figure 30.2 which describes the relation between the binding energy per nucleon and the number of nucleons we see that the lowest point is iron. This practically means that elements with lower atomic number than iron will participate in nuclear reactions where their mass increases. Elements with atomic numbers larger than iron will participate in nuclear reactions where the mass is lost. Therefore, we can expect that the heaviest element we could expect in old stars is iron.
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