Question

Explain the meaning of binding energy of a nucleus. In the nuclear reaction \(\ ^3Li^6 + \ ^0n^1 \longrightarrow \ ^2He^4 + \ ^1H^3 \), calculate the energy released in joules. Mass of \( \ ^3Li^6 = 6.015126~\text{u} \), Mass of \( \ ^2He^4 = 4.002604~\text{u} \), Mass of \( \ ^1H^3 = 3.016049~\text{u} \), Mass of \( \ ^0n^1 = 1.008665~\text{u} \), and 1u = 931 MeV.

Hint:

When calculating the energy released in a nuclear reaction, use the mass defect and convert it to energy using Einstein's equation: \(E = \Delta m . c^2\).

Answer 1

The binding energy of a nucleus is defined as the energy required to disassemble the nucleus into its constituent protons and neutrons. In simpler terms, it is the energy released when a nucleus is formed from its individual nucleons.
To calculate the energy released during the nuclear reaction, we use the mass-energy equivalence formula: \[ E = \Delta m . c^2 \] Where: - \( \Delta m \) is the mass defect (the difference between the mass of the reactants and products), - \( c \) is the speed of light (\( c = 3 \times 10^8 \, \text{m/s} \)). Step 1: Calculate the mass defect (\(\Delta m\)) The mass defect is the difference between the total mass of the reactants and the total mass of the products: \[ \Delta m = \left( \text{Mass of reactants} \right) - \left( \text{Mass of products} \right) \] For the reaction \( \ ^3Li^6 + \ ^0n^1 \longrightarrow \ ^2He^4 + \ ^1H^3 \), the mass defect is: \[ \Delta m = \left( 6.015126 + 1.008665 \right) - \left( 4.002604 + 3.016049 \right) \] \[ \Delta m = 7.023791 - 7.018653 = 0.005138~\text{u} \] Step 2: Convert the mass defect into energy To convert the mass defect into energy, we use the formula \( E = \Delta m . c^2 \). First, we need to convert the mass defect into kilograms: \[ 1~\text{u} = 1.660539 \times 10^{-27}~\text{kg} \] \[ \Delta m = 0.005138~\text{u} \times 1.660539 \times 10^{-27}~\text{kg/u} \] \[ \Delta m = 8.526 \times 10^{-30}~\text{kg} \] Now, applying Einstein's equation: \[ E = \Delta m . c^2 = 8.526 \times 10^{-30} . (3 \times 10^8)^2 \] \[ E = 8.526 \times 10^{-30} . 9 \times 10^{16} = 7.673 \times 10^{-13}~\text{J} \] So, the energy released in the reaction is: \[ E = 7.673 \times 10^{-13}~\text{J} \]