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Chapter 21: Problem 19

Calculate the nuclear binding energy (in J) and the nuclear binding energy per nucleon of the following isotopes: (a) \({ }_{3}^{7} \mathrm{Li}(7.01600 \mathrm{amu})\) and (b) 35 17 Cl \((34.95952 \mathrm{amu})\)

Short Answer

Expert verified
Total binding energy of \({ }_{3}^{7} \mathrm{Li}\) and \(^{35}_{17} \mathrm{Cl}\) can be calculated by following steps 1 to 4, and the binding energy per nucleon can be calculated in Step 5. The actual values will depend on the arithmetic performed in each step.

Step by step solution

01

Calculate the individual number of protons and neutrons

For \({ }_{3}^{7} \mathrm{Li}\), there are 3 protons and 4 neutrons (7 - 3). In \(^{35}_{17} \mathrm{Cl}\), there are 17 protons and 18 neutrons (35 - 17). Each proton has a mass of \(1.007825 \, amu\) and each neutron has a mass of \(1.008665 \, amu\).
02

Determine the Total Mass of Protons and Neutrons in Determined Nucleus for each isotope separately

In the case of \({ }_{3}^{7} \mathrm{Li}\), this would be \(3 \times 1.007825 + 4 \times 1.008665 \, amu\), and for \(^{35}_{17} \mathrm{Cl}\), it would be \(17 \times 1.007825 + 18 \times 1.008665 \, amu\)
03

Calculate the Mass Defect

The mass defect is the difference between the total mass calculated in Step 2 and the actual mass of the isotope given in the problem. For \({ }_{3}^{7} \mathrm{Li}\), subtract its actual mass (\(7.01600 \, amu\)) from the total mass calculated in Step 2. Do the same for \(^{35}_{17} \mathrm{Cl}\) using its actual mass (\(34.95952 \, amu\)).
04

Convert the Mass Defect of Each Isotope into Energy

One atomic mass unit (amu) is equivalent to \(931.5 \, MeV\). So, you can convert the mass defects calculated in Step 3 into energy by using the conversion factor.
05

Calculate the Nuclear Binding Energy per Nucleon

The nuclear binding energy per nucleon is the total nuclear binding energy divided by the number of nucleons. For \({ }_{3}^{7} \mathrm{Li}\), divide the total nuclear binding energy by 7, and for \(^{35}_{17} \mathrm{Cl}\), divide the total nuclear binding energy by 35.

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