Question

Using a mass spectrometer, the mass of the \({ }_{92}^{238} \mathrm{U}^{+}\) ion is found to be 238.05024 u. (a) Use this result to calculate the mass of the \({ }_{92}^{238} \mathrm{U}\) nucleus. (b) Now find the binding energy of the \({ }_{92}^{238} \mathrm{U}\) nucleus.

Short answer

Answer: The binding energy of the ${ }_{92}^{238} \mathrm{U}$ nucleus is approximately 169.128 MeV.

Step-by-step solution

(Part a: Mass of the Uranium nucleus)

First, let's find the mass of the \({ }_{92}^{238} \mathrm{U}\) nucleus. We are given that the mass of \({ }_{92}^{238} \mathrm{U}^{+}\) ion is 238.05024 u, which is the mass of the nucleus plus the mass of 91 electrons since one electron is removed to form the ion. Therefore, we can write: Mass of \({ }_{92}^{238} \mathrm{U}\) nucleus = Mass of \({ }_{92}^{238} \mathrm{U}^{+}\) ion - (Mass of 91 electrons) The mass of one electron is approximately 5.4858 x 10^{-4} u. So, Mass of \({ }_{92}^{238} \mathrm{U}\) nucleus = 238.05024 u - (91 * 5.4858 x 10^{-4} u) Calculating this we get, Mass of \({ }_{92}^{238} \mathrm{U}\) nucleus ≈ 237.94797 u

(Part b: Binding Energy of the Uranium nucleus)

To find the binding energy of the \({ }_{92}^{238} \mathrm{U}\) nucleus, we need to first calculate the mass defect. Mass defect is the difference between the mass of the nucleus and the mass of its constituent protons and neutrons. Mass defect = Mass of 92 protons + Mass of 146 neutrons - Mass of \({ }_{92}^{238} \mathrm{U}\) nucleus The mass of a proton is approximately 1.00728 u, and the mass of a neutron is approximately 1.00867 u. Mass defect = (92 * 1.00728 u) + (146 * 1.00867 u) - 237.94797 u Calculating this we get, Mass defect ≈ 0.18169 u Now we can find the binding energy using the mass-energy equivalence principle (E = mc^2) where E is the binding energy, m is the mass defect, and c is the speed of light. The speed of light c = 2.9979 x 10^8 m/s and 1 u = 931.494 MeV/c^2. Binding energy = Mass defect * c^2 * (Conversion factor) Binding energy = 0.18169 u * (2.9979 x 10^8 m/s)^2 * (931.494 MeV/c^2) Calculating this we get, Binding energy ≈ 169.128 MeV So, the binding energy of the \({ }_{92}^{238} \mathrm{U}\) nucleus is approximately 169.128 MeV.