Question

Explain the nuclear fusion process. What is its example in nature? If 4 neutrons and 3 protons are fused to form lithium \( {}_3^7 \text{Li} \) nucleus, then how much energy (in MeV) will be released?

Hint:

The energy released in a nuclear fusion reaction is directly related to the mass defect, and it can be calculated using Einstein’s equation \( E = \Delta m \cdot c^2 \).

Answer 1

Step 1: Nuclear Fusion Process.
Nuclear fusion is the process in which two light atomic nuclei combine to form a heavier nucleus, releasing a large amount of energy. This process is the primary source of energy in stars, including the Sun. An example of nuclear fusion in nature is the formation of helium from hydrogen in stars. In the Sun, four protons fuse to form a helium nucleus through a series of reactions, releasing a tremendous amount of energy in the process.
Step 2: Fusion of Neutrons and Protons.
In this case, 4 neutrons and 3 protons are fused to form a lithium nucleus (\( {}_3^7 \text{Li} \)).
Step 3: Mass Defect and Energy Released.
The energy released in a nuclear reaction is calculated using the mass defect and Einstein’s equation: \[ E = \Delta m \cdot c^2 \] where \( \Delta m \) is the mass defect (the difference between the total mass of the products and the mass of the reactants), and \( c \) is the speed of light (\( c = 3 \times 10^8 \, \text{m/s} \)). The mass of the particles involved in the reaction is: - Mass of proton = 1.00728 amu - Mass of neutron = 1.00867 amu - Mass of lithium nucleus \( {}_3^7 \text{Li} \) = 7.01436 amu
Step 4: Calculate Mass Defect.
The total mass of the reactants is: \[ \text{Mass of reactants} = (4 \times \text{Mass of neutron}) + (3 \times \text{Mass of proton}) = (4 \times 1.00867) + (3 \times 1.00728) = 4.03468 + 3.02184 = 7.05652 \, \text{amu} \] The mass of the product (lithium nucleus) is \( 7.01436 \, \text{amu} \). The mass defect is: \[ \Delta m = 7.05652 - 7.01436 = 0.04216 \, \text{amu} \]
Step 5: Convert Mass Defect to Energy.
To convert the mass defect to energy, we use the relation \( 1 \, \text{amu} = 931 \, \text{MeV} \): \[ E = 0.04216 \times 931 = 39.3 \, \text{MeV} \]
Step 6: Conclusion.
Thus, the energy released in the fusion process is \( \boxed{39.3 \, \text{MeV}} \).