CBSE · Class 12 · Physics · NucleiCalculate the energy released in the nuclear fusion reaction: $1^2H + 1^2H \rightarrow 2^3He + 0^1n$. \nGiven: \nMass of $1^2H = 2.014102 \text{ u}$\nMass of $2^3He = 3.016029 \text{ u}$\nMass of $0^1n = 1.008665 \text{ u}$\nAlso, $1 \text{ u} = 931.5 \text{ MeV}/c^2$. Show all steps clearly.

Step 1: Write down the given nuclear fusion reaction

$_1^2H + _1^2H \rightarrow _2^3He + _0^1n$

Step 2: Note the given masses of reactants and products

• Mass of deuterium reactant $m(_1^2H) = 2.014102 \text{ u}$

• Since there are two deuterium nuclei in the reactants, total reactant mass ($m_R$): $$m_R = 2 \times 2.014102 \text{ u} = 4.028204 \text{ u}$$

• Mass of helium-3 product $m(_2^3He) = 3.016029 \text{ u}$

• Mass of neutron product $m(_0^1n) = 1.008665 \text{ u}$

• Total product mass ($m_P$): $$m_P = 3.016029 \text{ u} + 1.008665 \text{ u} = 4.024694 \text{ u}$$

Step 3: Calculate the mass defect ($\Delta m$)

$$\Delta m = m_R - m_P$$ $$\Delta m = 4.028204 \text{ u} - 4.024694 \text{ u}$| $$\Delta m = 0.003510 \text{ u}$$

Step 4: Calculate the energy released ($E$)\nUsing the conversion factor $1 \text{ u} = 931.5 \text{ MeV}$, the energy released is:

$$E = \Delta m \times 931.5 \text{ MeV}$$ $$E = 0.003510 \times 931.5 \text{ MeV}$| $$E = 3.269565 \text{ MeV} \approx 3.27 \text{ MeV}$|

Conclusion\nThe total energy released in the given nuclear fusion reaction is approximately $3.27 \text{ MeV}$.