Question

Two deuterium nuclei, 2 1H, fuse to produce a helium nucleus, 3 2He , and a neutron. A neutral deuterium atom has a mass of 2.014102 u; a neutral helium atom has a mass of 3.016030 u; a neutral hydrogen atom has a mass of 1.007825 u; a neutron has a mass of 1.008665 u; and a proton has a mass of 1.007277 u. How much energy is released in the process? 1 u = 931.494 MeV/c2.

Answer 1

The energy released in the process is 3.27 MeV or 5.23 x 10⁻¹³ J

Step-by-step explanation:

For the release of energy there must be a difference in energies of the reactants and products. The released energy will be equal to this difference in energies of reactants and products.

The difference in mass of reactant and product is:

Difference in mass = Δm = Mass of Reactant - Mass of Products

Δm = 2(Mass of Deuterium) - Mass of Helium - Mass of Neutron

Δm = 2(2.014102 u) - 3.016030 u - 1.008665 u

Δm = 4.028204 u - 3.016030 u - 1.008665 u

Δm = 0.003509 u

This difference in mass corresponds to the released energy. In order to convert this to energy, we use relation:

Released Energy = ΔE = Δm * 931.494 MeV/u

ΔE = (0.003509 u)(931.494 MeV/u)

ΔE = 3.27 MeV

Now, converting it to Joules:

ΔE = (3.27 x 10⁶ eV)(1.6 x 10⁻¹⁹ J/1 eV)

ΔE = 5.23 x 10⁻¹³ J