Question

In the nuclear reaction:

\[\ce{^2_1H + ^2_1H -> ^A_ZX + ^1_0n}\]

1. Find the value of A.
2. Calculate the amount of energy released in the reaction.

Given: \[\ce{m(^2_1H)}\] = 2.014102 u, \[\ce{m(^A_ZX)}\] = 3.016049 u, \[\ce{m(^1_0n)}\] = 1.008665 u, 1 u = 931.5 MeV/c²

Answer 1

In a nuclear reaction, mass number (A) and atomic number (Z) are conserved.

The energy released (Q-value) is due to the mass defect (∆m) converted into energy using:

E = ∆m - c²

Conservation of A:

`sum A_"reactants" = sum A_"products"`

Mass defect ∆m:

`sum m_"reactants" - sum m_"products"`

Q = ∆m(in u) × 931.5 MeV/u

a. Value of A:

Conservation of mass number:

2 + 2 = A + 1

⇒  A = 3

Conservation of atomic number:

1 + 1 = Z + 0

⇒ Z = 2

Thus, the product X is Helium-3 \[\ce{^3_2 He}\].

b. Energy Released:

Total mass of reactants (M_R) = 2 × 2.014102

= 4.028204 u

Total mass of products (M_P) = 3.016049 + 1.008665

= 4.024714 u

Mass defect (∆m) = M_R − M_P

= 4.028204 − 4.024714

= 0.00349 u

Energy released (Q) = 0.00349 × 931.5

≈ 3.2509 MeV