Question

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

The nucleus with the largest binding energy per nucleon is \({ }^{9} \mathrm{Be}\), which is calculated using the mass defects, mass-energy equivalence equation, and dividing the binding energy by the number of nucleons in each nucleus.

Step-by-step solution

List known quantities

We are given the nuclear masses of \({ }^{7} \mathrm{Be}\), \({ }^{9} \mathrm{Be}\), and \({ }^{10} \mathrm{Be}\), which are: \(M({ }^{7} \mathrm{Be}) = 7.0147\,\text{amu}\) \(M({ }^{9} \mathrm{Be}) = 9.0100\,\text{amu}\) \(M({ }^{10} \mathrm{Be}) = 10.0113\,\text{amu}\) We also know that: 1 amu = \(1.66054\times 10^{-27}\,\) kg 1 MeV = \(1.60219\times 10^{-13}\,\) J Mass of a neutron = 1.0087 amu Mass of a proton = 1.0073 amu

Calculate the mass defect for each nucleus

For each nucleus, we will find the mass defect by subtracting the sum of the individual neutron and proton masses from the total atomic mass. For \({ }^{7} \mathrm{Be}\): Mass defect = \([4M_p + 3M_n - M({ }^{7} \mathrm{Be})]\) = \([4 \times 1.0073 + 3 \times 1.0087 - 7.0147]\,\text{amu}\) For \({ }^{9} \mathrm{Be}\): Mass defect = \([5M_p + 4M_n - M({ }^{9} \mathrm{Be})]\) = \([5 \times 1.0073 + 4 \times 1.0087 - 9.0100]\,\text{amu}\) For \({ }^{10} \mathrm{Be}\): Mass defect = \([5M_p + 5M_n - M({ }^{10} \mathrm{Be})]\) = \([5 \times 1.0073 + 5 \times 1.0087 - 10.0113]\,\text{amu}\)

Convert mass defect to energy

Now we have the mass defect in amu for each nucleus. Next, convert each mass defect to energy using Einstein's mass-energy equivalence equation: \(E = mc^2\) First, convert each mass defect in amu to its equivalent in kg by multiplying with the conversion factor: \(Mass\, Defect\,(kg) = Mass\, Defect\,(amu) \times 1.66054\times 10^{-27}\,\text{kg/amu}\) Next, calculate the energy in Joules by using the mass-energy equivalence equation: \(Energy\,(J) = Mass\, Defect\,(kg) \times (3\times 10^8\,\text{m/s})^2\) Now, convert the energy in Joules to MeV using the conversion factor: \(Energy\,(MeV) = Energy\,(J) \times \frac{1}{1.60219\times 10^{-13}\,\text{J/MeV}}\)

Calculate binding energy per nucleon

For each nucleus, divide the binding energy obtained in Step 3 by the number of nucleons in that nucleus to get the binding energy per nucleon. For \({ }^{7} \mathrm{Be}\): BE per nucleon = \(\frac{Binding\,Energy\,of\,{}^{7} \mathrm{Be}}{7}\) For \({ }^{9} \mathrm{Be}\): BE per nucleon = \(\frac{Binding\,Energy\,of\,{}^{9} \mathrm{Be}}{9}\) For \({ }^{10} \mathrm{Be}\): BE per nucleon = \(\frac{Binding\,Energy\,of\,{}^{10} \mathrm{Be}}{10}\)

Determine which nucleus has the largest binding energy per nucleon

Now that we have the binding energy per nucleon for each nucleus, we can compare them and determine which nucleus has the largest binding energy per nucleon. The nucleus with the highest value represents the largest binding energy per nucleon.