Question

What is the difference between nuclear fusion and nuclear fission? Find the value of the energy \( Q \) released with the help of the given nuclear fusion reaction:
\[ \text{1} \, ^1\text{H}_2 + \, ^1\text{H}_3 ⇒ \, ^2\text{He}_4 + \, ^1\text{n}^1 + Q \] Given:
Mass of \( ^1\text{H}_2 = 2.0141 \, \text{amu} \)
Mass of \( ^1\text{H}_3 = 3.0160 \, \text{amu} \)
Mass of \( ^2\text{He}_4 = 4.0026 \, \text{amu} \)
Mass of \( ^1\text{n}^1 = 1.0087 \, \text{amu} \)
1 amu = 931 MeV

Hint:

The energy released in a nuclear reaction is directly related to the mass defect. For fusion and fission, the energy can be calculated by finding the mass difference between the reactants and products and multiplying by 931 MeV per amu.

Answer 1

Nuclear Fusion and Nuclear Fission are two types of nuclear reactions:
- Nuclear Fusion is the process in which two light nuclei combine to form a heavier nucleus, releasing a large amount of energy. For example, in stars, hydrogen nuclei fuse to form helium, releasing energy.
- Nuclear Fission is the process in which a heavy nucleus splits into two or more lighter nuclei, accompanied by the release of energy. Fission is commonly used in nuclear reactors.
Energy Released in Nuclear Fusion Reaction: The energy released in a nuclear reaction is given by the difference in mass between the reactants and products, multiplied by \( c^2 \) (where \( c \) is the speed of light). The mass defect (\( \Delta m \)) is the difference between the total mass of the reactants and the total mass of the products.
Step 1: Calculate the mass defect.
The total mass of the reactants is the sum of the masses of \( ^1\text{H}_2 \) and \( ^1\text{H}_3 \):
\[ m_{\text{reactants}} = m_{^1\text{H}_2} + m_{^1\text{H}_3} = 2.0141 + 3.0160 = 5.0301 \, \text{amu} \] The total mass of the products is the sum of the masses of \( ^2\text{He}_4 \) and \( ^1\text{n}^1 \):
\[ m_{\text{products}} = m_{^2\text{He}_4} + m_{^1\text{n}^1} = 4.0026 + 1.0087 = 5.0113 \, \text{amu} \] The mass defect \( \Delta m \) is:
\[ \Delta m = m_{\text{reactants}} - m_{\text{products}} = 5.0301 - 5.0113 = 0.0188 \, \text{amu} \] Step 2: Convert mass defect to energy.
Using the equivalence of mass and energy (\( E = \Delta m \times c^2 \)), and knowing that \( 1 \, \text{amu} = 931 \, \text{MeV}/c^2 \), the energy released \( Q \) is:
\[ Q = \Delta m \times 931 = 0.0188 \times 931 = 17.5 \, \text{MeV} \] Thus, the energy released in this nuclear fusion reaction is \( Q = 17.5 \, \text{MeV} \).