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Chapter 21: Problem 79

The nuclear masses of \({ }^{7} \mathrm{Be},{ }^{9} \mathrm{Be},\) and \({ }^{10} \mathrm{Be}\) are \(7.0147,9.0100,\) and 10.0113 amu, respectively. Which of these nuclei has the largest binding energy per nucleon?

Short Answer

Expert verified
The isotope with the largest binding energy per nucleon is \({ }^{9}\)Be, with a binding energy of approximately 4.809 MeV/nucleon.

Step by step solution

01

Recall the formula for binding energy of a nucleus

The binding energy of a nucleus is given by the mass defect multiplied by the speed of light squared: \[ E = \Delta m \times c^2 \] where \(\Delta m\) is the mass defect and \(c\) is the speed of light.
02

Calculate the mass defect for each isotope

To calculate the mass defect, we will subtract the nuclear mass from the combined mass of all the protons and neutrons present in the nucleus. For beryllium, we know that it has 4 protons and the number of neutrons varies depending on the isotope. The mass of a single proton is 1.00728 amu and the mass of a single neutron is 1.00867 amu. For \({ }^{7}\)Be: \[ \Delta m = (4 \times 1.00728) + (3 \times 1.00867) - 7.0147 = 0.02919 \text{ amu} \] For \({ }^{9}\)Be: \[ \Delta m = (4 \times 1.00728) + (5 \times 1.00867) - 9.0100 = 0.04647 \text{ amu} \] For \({ }^{10}\)Be: \[ \Delta m = (4 \times 1.00728) + (6 \times 1.00867) - 10.0113 = 0.03939 \text{ amu} \]
03

Calculate the binding energy for each isotope

Now that we have the mass defects, we can use the formula for binding energy to find the binding energy for each isotope. We will use the conversion factor \(1 \text{ amu} = 931.5 \text{ MeV/c}^2\) to convert the mass defects into MeV. For \({ }^{7}\)Be: \[ E = 0.02919 \times 931.5 = 27.179 \text{ MeV} \] For \({ }^{9}\)Be: \[ E = 0.04647 \times 931.5 = 43.281 \text{ MeV} \] For \({ }^{10}\)Be: \[ E = 0.03939 \times 931.5 = 36.696 \text{ MeV} \]
04

Calculate the binding energy per nucleon for each isotope

Finally, we will divide the binding energy by the number of nucleons in each isotope. For \({ }^{7}\)Be: \[ \frac{E}{A} = \frac{27.179}{7} = 3.883 \text{ MeV/nucleon} \] For \({ }^{9}\)Be: \[ \frac{E}{A} = \frac{43.281}{9} = 4.809 \text{ MeV/nucleon} \] For \({ }^{10}\)Be: \[ \frac{E}{A} = \frac{36.696}{10} = 3.670 \text{ MeV/nucleon} \]
05

Identify the isotope with the largest binding energy per nucleon

Comparing the binding energy per nucleon for each isotope: \[ { }^{7} \mathrm{Be}: 3.883 \text{ MeV/nucleon}\\ { }^{9} \mathrm{Be}: 4.809 \text{ MeV/nucleon}\\ { }^{10} \mathrm{Be}: 3.670 \text{ MeV/nucleon} \] We can see that \({ }^{9}\)Be has the largest binding energy per nucleon, which is approximately 4.809 MeV/nucleon.

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