What is the energy associated with the formation of 2.55 g of 4He by the fusion of 3H and 1H?

Substance Mass (u)
4He 4.00260
3H 3.01605
1H 1.00783

The energy associated with the formation of 2.55 g of 4He by the fusion of 3H and 1H is -2.982 x 10⁻¹⁰ J.

The given masses of the isotopes can be converted to kilograms using the conversion factor: 1 u = 1.661 x 10⁻²⁷ kg.

Mass of 4He = 2.55 g = 2.55 x 10⁻³ kg

Mass of 3H = 3.01605 u = 3.01605 x 1.661 x 10⁻²⁷ kg/u

= 5.0099 x 10⁻²⁷ kg

Mass of 1H = 1.00783 u = 1.00783 x 1.661 x 10⁻²⁷ kg/u

= 1.6737 x 10⁻²⁷ kg

The balanced equation for the fusion reaction is;

3H + 1H → 4He

The molar mass of 4He is 4.0026 g/mol, which can be converted to kg/mol using the conversion factor: 1 g/mol = 1 x 10⁻³ kg/mol.

Molar mass of 4He = 4.0026 g/mol = 4.0026 x 10⁻³ kg/mol

The number of moles of 4He formed can be calculated from its mass;

n(4He) = m(4He) / M(4He)

= 2.55 x 10⁻³ kg / 4.0026 x 10⁻³ kg/mol

= 0.638 mol

From the balanced equation, 3 moles of H atoms react with 1 mole of He atoms to form 1 mole of He atoms. Therefore, the number of moles of H atoms required for the reaction is;

n(H) = 3/4 x n(4He)

= 3/4 x 0.638 mol

= 0.479 mol

The energy released in the reaction can be calculated using the mass-energy equivalence equation;

E = Δm c²

where Δm is change in mass, c is the speed of light.

The change in mass is;

Δm = [3H + 1H - 4He] = [5.0099 x 10⁻²⁷ kg + 1.6737 x 10⁻²⁷kg - 4.0026 x 10⁻³ kg]

= -3.315 x 10⁻²⁷ kg (negative because mass is lost in the reaction)

The energy released is;

E = (-3.315 x 10⁻²⁷ kg) c²

= (-3.315 x 10⁻²⁷ kg) (2.998 x 10⁸ m/s)²

= -2.982 x 10⁻¹⁰ J

The negative sign indicates that energy is released in the reaction (exothermic reaction).

Therefore, the energy associated is -2.982 x 10⁻¹⁰ J.

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