Problem

Find the energy released in the following fusion reaction:

$${}_{1}^{2}\mathrm{H} + {}_{1}^{3}\mathrm{H} \rightarrow {}_{2}^{4}\mathrm{He} + {}_{0}^{1}\mathrm{n}$$

Data

• Initial masses:
  • Deuterium (${}_{1}^{2}\mathrm{H}$): $2.014 \, \mathrm{u}$
  • Tritium (${}_{1}^{3}\mathrm{H}$): $3.016 \, \mathrm{u}$
  • Total initial mass:

$$m_{\text{initial}} = 5.0302 \, \mathrm{u}$$

• Final masses:
  • Helium (${}_{2}^{4}\mathrm{He}$): $4.0026 \, \mathrm{u}$
  • Neutron (${}_{0}^{1}\mathrm{n}$): $1.0087 \, \mathrm{u}$
  • Total final mass:

$$m_{\text{final}} = 5.0113 \, \mathrm{u}$$

• Mass-energy equivalence: $1 \, \mathrm{u} = 931.5 \, \mathrm{MeV}$

Prerequisite Concepts

1. Mass Defect ($\Delta m$):
   The difference between the total initial mass and the total final mass.

$$\Delta m = m_{\text{initial}} - m_{\text{final}}$$

2. Energy Released ($Q$):
   The energy released is calculated using the mass-energy equivalence relation:

$$Q = \Delta m \cdot 931.5 \, \mathrm{MeV}$$

Solution

Step 1: Calculate the Mass Defect ($\Delta m$)

Using the formula:

$$\Delta m = m_{\text{initial}} - m_{\text{final}}$$

Substitute the given values:

$$\Delta m = 5.0302 \, \mathrm{u} - 5.0113 \, \mathrm{u}$$ $$\Delta m = 0.0189 \, \mathrm{u}$$

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Step 2: Calculate the Energy Released ($Q$)

Using the formula:

$$Q = \Delta m \cdot 931.5 \, \mathrm{MeV}$$

Substitute $\Delta m = 0.0189 \, \mathrm{u}$:

$$Q = 0.0189 \cdot 931.5$$ $$Q = 17.6 \, \mathrm{MeV}$$

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Answer

The energy released in the given fusion reaction is:

$$Q = 17.6 \, \mathrm{MeV}$$