Question

In the fusion reaction,

The masses of deuteron, helium and neutron are 2.015 u, 0.017 u and 1.009 u respectively. If 1 kg \(_{1}^{2} {H}+_{1}^{2} {H} \longrightarrow_{2}^{3} {H} {e}+_{0}^{1} n\)of deuteron completely undergoes fusion, then calculate free energy.
[1u = 9.31 MeV/c² ]

Answer 1

There are 6.02 × 10²³ atoms in 1 mole (0.002 kg) of deuterium.
∴ Number of atoms in 1 kg of deuterium
= \(\frac{6.023 \times 10^{23}}{0.002}\)
= 3.01 × 10²⁶
From above equation, for produced energy from fusion of deuterium
Δm = [2 × 2.015 u – (3.017 + 1.009) u]
Δm = [4.030u – 4.026u]
Δm = 0.004
E_B = Δmc²
E_B = 0.004 × 931 × \(\frac { { Me }{ V } }{ c^{ 2 } } c^{ 2 }\)
E_B = 3.724 Me V
Fusion energy of 2 deuterium = 3.724 MeV
Fusion energy of 1 deuterium = \(\frac{3.724}{2}\) Me V = 1.862 Me V
Fusion energy of 1 kg deuterium
= (1.862 × 3.01 × 10²⁶) MeV
= 5.6 × 10²⁶ MeV
= 5.6 × 10²⁶ × 10⁶ × 1.6 × 10^-19J
= 8.96 × 10¹³
= 9 × 10¹³ J