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Which has the larger binding energy, \(\mathrm{K}-40\) or \(\mathrm{Ca}-40 ?\) Show by calculation.

Short Answer

Expert verified
Potassium-40 (K-40) has a larger binding energy compared to Calcium-40 (Ca-40). The binding energy for K-40 is 215.07 MeV, while the binding energy for Ca-40 is 183.11 MeV.

Step by step solution

01

Find the mass of the constituent protons, neutrons, and the nucleus of K-40

Look up the mass of a proton, neutron, and the mass of \(\mathrm{K}-40\) isotope (in atomic mass units, amu). Then, determine the number of protons and neutrons in the nucleus. Here's the mass of a proton, neutron, and the nucleus of K-40: - Mass of a proton: \(m_p = 1.0073\: \text{amu}\) - Mass of a neutron: \(m_n = 1.0087\: \text{amu}\) - Mass of \(\mathrm{K}-40\) nucleus: \(m_\text{K-40} = 39.963998\:\text{amu}\) Now, let's find the number of protons and neutrons in K-40 nucleus: - Number of protons \((Z) = 19\) - Number of neutrons \((N) = 21\)
02

Calculate the mass defect of K-40

Subtract the mass of the nucleus from the mass of protons and neutrons combined: \(\Delta m_\text{K-40}(19\cdot m_p + 21\cdot m_n) - m_\text{K-40} = (19\cdot 1.0073\: \text{amu} + 21\cdot 1.0087\: \text{amu}) - 39.963998\: \text{amu} = 0.230892\: \text{amu}\)
03

Calculate the binding energy of K-40

Multiply the mass defect by the speed of light squared. We will use the value of \(c^2 = 931.5\: \text{MeV}/\text{amu}\) to do the conversion from atomic mass units to MeV (mega-electron-volts): \(BE_\text{K-40} = \Delta m_\text{K-40} * c^2 = 0.230892\: \text{amu} * 931.5\: \text{MeV}/\text{amu} = 215.07\: \text{MeV}\) Now, let's do the same for Calcium-40 (\(\mathrm{Ca}-40\)):
04

Find the mass of the nucleus of Ca-40

Look up the mass of \(\mathrm{Ca}-40\) isotope (in atomic mass units, amu). Then, determine the number of protons and neutrons in the nucleus. Here's the mass of the nucleus of Ca-40: - Mass of \(\mathrm{Ca}-40\) nucleus: \(m_\text{Ca-40} = 39.962591\: \text{amu}\) Now, let's find the number of protons and neutrons in Ca-40 nucleus: - Number of protons \((Z) = 20\) - Number of neutrons \((N) = 20\)
05

Calculate the mass defect of Ca-40

Subtract the mass of the nucleus from the mass of protons and neutrons combined: \(\Delta m_\text{Ca-40}(20\cdot m_p + 20\cdot m_n) - m_\text{Ca-40} = (20\cdot 1.0073\: \text{amu} + 20\cdot 1.0087\: \text{amu}) - 39.962591\: \text{amu} = 0.1965\: \text{amu}\)
06

Calculate the binding energy of Ca-40

Multiply the mass defect by the speed of light squared: \(BE_\text{Ca-40} = \Delta m_\text{Ca-40} * c^2 = 0.1965\: \text{amu} * 931.5\: \text{MeV}/\text{amu} = 183.11\: \text{MeV}\) Now we can compare the binding energies of both isotopes: \(BE_\text{K-40} = 215.07\: \text{MeV}\) and \(BE_\text{Ca-40} = 183.11\: \text{MeV}\) Therefore, Potassium-40 (\(\mathrm{K}-40\)) has a larger binding energy compared to Calcium-40 (\(\mathrm{Ca}-40\)).

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