CBSE · Class 12 · Physics · NucleiCalculate the energy released in MeV in the nuclear reaction: ${1}^{2}\text{H} + {1}^{3}\text{H} \rightarrow {2}^{4}\text{He} + {0}^{1}\text{n}$\nGiven masses:\nMass of ${1}^{2}\text{H} = 2.014102 \text{ u}$\nMass of ${1}^{3}\text{H} = 3.016049 \text{ u}$\nMass of ${2}^{4}\text{He} = 4.001506 \text{ u}$\nMass of ${0}^{1}\text{n} = 1.008665 \text{ u}$ (Use $1 \text{ u} = 931.5 \text{ MeV}$)

Step 1: Write down the given nuclear reaction

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

Step 2: Calculate the total mass of the reactants

$$\text{Mass of } _{1}^{2}\text{H} = 2.014102 \text{ u}$$ $$\text{Mass of } _{1}^{3}\text{H} = 3.016049 \text{ u}$$ $$\text{Total mass of reactants } (m_R) = 2.014102 + 3.016049 = 5.030151 \text{ u}$$

Step 3: Calculate the total mass of the products

$$\text{Mass of } _{2}^{4}\text{He} = 4.001506 \text{ u}$$ $$\text{Mass of } _{0}^{1}\text{n} = 1.008665 \text{ u}$$ $$\text{Total mass of products } (m_P) = 4.001506 + 1.008665 = 5.010171 \text{ u}$$

Step 4: Calculate the mass defect ($\Delta m$)

$$\Delta m = m_R - m_P$$ $$\Delta m = 5.030151 \text{ u} - 5.010171 \text{ u} = 0.019980 \text{ u}$$

Step 5: Convert the mass defect into energy\nSince $1 \text{ u} \approx 931.5 \text{ MeV}$, the energy released $Q$ is:

$$Q = \Delta m \times 931.5 \text{ MeV}$$ $$Q = 0.019980 \times 931.5 = 18.61177 \text{ MeV}$$

Conclusion\nThe energy released in the given fusion reaction is approximately 18.61 MeV.