Question

(a) Consider the so-called ‘D-T reaction’ (Deuterium-Tritium reaction).
In a thermonuclear fusion reactor, the following nuclear reaction occurs: \[ \ ^{2}_1 \text{H} + \ ^{3}_1 \text{H} \longrightarrow \ ^{4}_2 \text{He} + \ ^{1}_0 \text{n} + Q \] Find the amount of energy released in the reaction.
% Given data Given:
\( m\left(^{2}_1 \text{H}\right) = 2.014102 \, \text{u} \)
\( m\left(^{3}_1 \text{H}\right) = 3.016049 \, \text{u} \)
\( m\left(^{4}_2 \text{He}\right) = 4.002603 \, \text{u} \)
\( m\left(^{1}_0 \text{n}\right) = 1.008665 \, \text{u} \)
\( 1 \, \text{u} = 931 \, \text{MeV}/c^2 \)

Hint:

The energy released in nuclear reactions is a result of the conversion of mass into energy. This is why mass defect (difference between reactants and products) is multiplied by \( c^2 \) to find the energy released.

Answer 1

The energy output from a nuclear reaction is calculated by multiplying the mass defect (the difference between the total mass of the reactants and the total mass of the products) by the square of the speed of light: \[ Q = (\Delta m) c^2 \] where: \[ \Delta m = \left( m_{\text{reactants}} - m_{\text{products}} \right) \] For this specific reaction: \[ m_{\text{reactants}} = m\left(^{2}_1 \text{H}\right) + m\left(^{3}_1 \text{H}\right) = 2.014102 \, \text{u} + 3.016049 \, \text{u} = 5.030151 \, \text{u} \] \[ m_{\text{products}} = m\left(^{4}_2 \text{He}\right) + m\left(^{1}_0 \text{n}\right) = 4.002603 \, \text{u} + 1.008665 \, \text{u} = 5.011268 \, \text{u} \] The resulting mass defect is: \[ \Delta m = m_{\text{reactants}} - m_{\text{products}} = 5.030151 \, \text{u} - 5.011268 \, \text{u} = 0.018883 \, \text{u} \] Converting the mass defect to energy: \[ Q = \Delta m \cdot c^2 = 0.018883 \, \text{u} \times 931 \, \text{MeV}/c^2 \] \[ Q = 17.6 \, \text{MeV} \] The energy released in this reaction is: \[ \boxed{Q = 17.6 \, \text{MeV}} \]