Question

Consider the following nuclear fusion reaction

$\qquad ^{2}_{1}H + ^{2}_{1}H \implies ^{3}_{1}H + ^{1}_{1}P$

Assume binding energy per nucleon of deuterium and tritium are 1 MeV and 2.80 MeV respectively. Neglect the kinetic energy of the nuclei before the fusion. What will be the kinetic energy (in MeV) of the tritium produced in the above nuclear reaction.

• A. 1.1MeV
• B. 2.2MeV
• C. 3.3MeV
• D. 2.8MeV

Answer 1

Answer: (A)

1.1MeV

Answer 2

The nuclear fusion reaction given is:

$\qquad ^{2}_{1}H + ^{2}_{1}H \implies ^{3}_{1}H + ^{1}_{1}P$

First, we calculate the total binding energy of the reactants and products.
The binding energy (BE) of a nucleus is given by (Binding Energy per Nucleon) × (Number of Nucleons).

1. Binding Energy of Reactants:

   • Deuterium ($^{2}_{1}H$) has 2 nucleons.
   • Binding energy per nucleon of deuterium = 1 MeV.
   • Total binding energy of one deuterium nucleus = $1 \text{ MeV/nucleon} \times 2 \text{ nucleons} = 2 \text{ MeV}$.
   • Since there are two deuterium nuclei reacting, the total binding energy of reactants = $2 \times 2 \text{ MeV} = 4 \text{ MeV}$.

2. Binding Energy of Products:

   • Tritium ($^{3}_{1}H$) has 3 nucleons.
   • Binding energy per nucleon of tritium = 2.80 MeV.
   • Total binding energy of one tritium nucleus = $2.80 \text{ MeV/nucleon} \times 3 \text{ nucleons} = 8.40 \text{ MeV}$.
   • A proton ($^{1}_{1}P$) is a single nucleon and does not have binding energy in this context (its binding energy is considered zero).
   • Total binding energy of products = $8.40 \text{ MeV} + 0 \text{ MeV} = 8.40 \text{ MeV}$.

3. Q-value (Energy Released) of the Reaction:
   The energy released in a nuclear reaction (Q-value) is the difference between the total binding energy of the products and the total binding energy of the reactants.
   $Q = \text{Total Binding Energy of Products} - \text{Total Binding Energy of Reactants}$
   $Q = 8.40 \text{ MeV} - 4 \text{ MeV} = 4.40 \text{ MeV}$
   This 4.40 MeV is released as kinetic energy of the products (tritium and proton).

4. Kinetic Energy Distribution among Products:
   Since the kinetic energy of the nuclei before fusion is neglected, the initial momentum of the system is zero. By the principle of conservation of momentum, the total momentum of the products after the reaction must also be zero.
   Let $m_T$ and $m_P$ be the masses of tritium and proton, respectively.
   Let $v_T$ and $v_P$ be their velocities.
   $m_T v_T + m_P v_P = 0$
   This implies that the magnitudes of their momenta are equal: $|p_T| = |p_P| = p$.

   The kinetic energy ($K$) of a particle with momentum $p$ and mass $m$ is given by $K = \frac{p^2}{2m}$.
   So, for tritium: $K_T = \frac{p^2}{2m_T}$
   And for proton: $K_P = \frac{p^2}{2m_P}$
   From these equations, we can see that $2m_T K_T = p^2$ and $2m_P K_P = p^2$.
   Therefore, $m_T K_T = m_P K_P$.

   We can approximate the masses of the nuclei by their mass numbers (A):
   $m_T \approx A_T = 3$ (for tritium)
   $m_P \approx A_P = 1$ (for proton)
   So, $3 K_T = 1 K_P \implies K_P = 3 K_T$.

   The total kinetic energy released (Q-value) is shared between the products:
   $Q = K_T + K_P$
   Substitute $K_P = 3 K_T$ into this equation:
   $4.40 \text{ MeV} = K_T + 3 K_T$
   $4.40 \text{ MeV} = 4 K_T$
   $K_T = \frac{4.40 \text{ MeV}}{4}$
   $K_T = 1.1 \text{ MeV}$

The kinetic energy of the tritium produced in the reaction is 1.1 MeV.