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When hydrogen nuclei are fused together to form a helium nucleus, the mass of the product is 0.7% less than the mass of the reactants. If 1 gram (1 × 10-3 kg) of protons were completely fused into helium nuclei, the mass of the helium nuclei produced would be 6.8 × 10-6 kg less than the mass of the protons. Using the equation E = mc2, calculate how many joules of energy would be produced by converting 6.8 × 10-6 kg of matter into energy in this fusion reaction. Use c = 3 × 108 m/s.

Respuesta :

sebasacevedop

Answer:

[tex]E=6.12*10^{11}J[/tex]

Explanation:

As noted, this formula states that anything that has mass contains a corresponding energy, given by its mass multiplied by the speed of light squared. Similarly, energy contains a corresponding mass given by its energy divided by the speed of light squared.

So, replacing in the formula:

[tex]E=6.8*10^{-6}kg(3*10^8\frac{m}{s})^2\\E=6.12*10^{11}J[/tex]

deniseerwin

Answer:

6.12 × 1011 kg-m2/s2

Explanation:

Answer for Plato

The energy can be calculated as shown.

E = mc2

  = (6.8 × 10-6 kg) × (3 × 108 m/s)2

  = (6.8 × 10-6 kg) × (9 × 1016 m2/s2)

  = 61.2 × 1010 kg-m2/s2

  = 6.12 × 1011 kg-m2/s2

So, the amount of energy created by fusing 1 gram of protons is 6.12 × 1011 kg-m2/s2.

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