Calculate the binding energy for \(_{1}^{2} \mathrm{H}\) and $_{1}^{3} \mathrm{H} .$ The atomic masses are \(_{1}^{2} \mathrm{H}, 2.01410 \mathrm{u} ;\) and $^{3} \mathrm{H}, 3.01605 \mathrm{u} .$

Short Answer

Expert verified
The binding energy for \(_{1}^{2} \mathrm{H}\) is \(4.632*10^{-14} J\) and for \(_{1}^{3} \mathrm{H}\) is \(2.135*10^{-13} J\).

Step by step solution

01

Calculate the mass defect for \(_{1}^{2} \mathrm{H}\) and \(_{1}^{3} \mathrm{H}\)

Find the difference between the masses of the protons and neutrons in the nucleus and the actual mass of each isotope. The mass of a proton is approximately 1.00728 u and the mass of a neutron is approximately 1.00867 u. - For \(_{1}^{2} \mathrm{H}\): Mass of 1 proton + Mass of 1 neutron - Mass of \(_{1}^{2} \mathrm{H}\) - For \(_{1}^{*>}<*>3>\mathrm{H}\): Mass of 1 proton + 2 * Mass of 1 neutron - Mass of \(_{1}^{3} \mathrm{H}\)
02

Calculate the mass defect for each isotope

- Mass defect of \(_{1}^{2} \mathrm{H}\): \(\Delta m_1 = 1.00728 \mathrm{u} + 1.00867 \mathrm{u} - 2.01410 \mathrm{u} \Rightarrow \Delta m_1 = 0.00185 \mathrm{u}\) - Mass defect of \(_{1}^{3} \mathrm{H}\): \(\Delta m_2 = 1.00728 \mathrm{u} + 2 * 1.00867 \mathrm{u} - 3.01605 \mathrm{u} \Rightarrow \Delta m_2 = 0.00857 \mathrm{u}\)
03

Convert mass defect to binding energy

We will now use Einstein's equation to convert the mass defect (\(\Delta m\)) into energy. The equation is \(E=mc^2\) where \(E\) is the energy, \(m\) is the mass, and \(c\) is the speed of light. The speed of light, \(c = 2.998 * 10^8 ms^{-1}\). We need to convert the mass defect to kg first. We know that \(1 \mathrm{u} = 1.66054 * 10^{-27} kg\). Then we will find the binding energy for each isotope by multiplying their mass defect by \(c^2\).
04

Calculate the binding energy for each isotope

- Binding energy of \(_{1}^{2} \mathrm{H}\): \(E_1 = \Delta m_1 * c^2 \Rightarrow E_1 = (0.00185 \mathrm{u}) * (1.66054 * 10^{-27} kg/u) * (2.998 * 10^8 m/s)^2 \Rightarrow E_1 = 4.632 *10^{-14} J\) - Binding energy of \(_{1}^{3} \mathrm{H}\): \(E_2 = \Delta m_2 * c^2 \Rightarrow E_2 = (0.00857 \mathrm{u}) * (1.66054 * 10^{-27} kg/u) * (2.998 * 10^8 m/s)^2 \Rightarrow E_2 = 2.135 *10^{-13} J\) The binding energy for \(_{1}^{2} \mathrm{H}\) is \(4.632*10^{-14} J\) and for \(_{1}^{3} \mathrm{H}\) is \(2.135*10^{-13} J\).

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