The mass loss experienced by the fusion of the nucleons is due to the nuclear binding energy in the nucleus.

The mass loss is (1.007825 + 1.008660) - 2.014100 = 0.002390 atomic mass units

Convert the atomic mass units into grams.

0.002390/6.023 X 10²³ = 3.97 X 10^-27 grams or 3.97 X 10^-30kilograms
Using the formula E = mc² find the nuclear binding energy

E = mc²

E = 3.97 X 10^-30 X (3.00 X 10⁸)²

E = 3.57 X 10^-13J

 

2) Calculate the mole of deuteron in 6 grams of the substance.

6/2 = 3 mole of deuteron nuclei.
Since 3.57 X 10^-13J of energy is the nuclear binding energy per deuteron the nuclear binding energy in 6 grams of deuteron is given by the expression

3 X 6.023 X 10²³ X 3.57 X 10^-13 = 6.26 X 10¹¹J

 

3)a) Write the nuclear equation that represents the formation of a helium nucleus from the fusion of four protons

.b) Given that 4 protons fuse to form a helium nucleus of relative mass 4.002604 atomic mass units and a positron of relative mass 0.000549 atomic mass units, calculate the amount of energy that is given out when 12 grams of helium is formed.

Calculate the mass loss for the formation of the helium nucleus.

4 X protons - (helium + 2 X positrons)
4 X 1.007825 - (4.002604 + 2 X 0.000549) =0.027598 atomic mass units.

Calculate the mass loss, in grams, per helium nucleus
0.027598 / 6.023 X 10²³ = 4.58 X 10^-26 grams or 4.58 X 10^-29 kilograms

Calculate the energy per helium nucleus

E = mc²

E = 4.58 X 10^-29 X (3.00 X 10⁸)²

E = 4.122 X 10^-12J

Calculate the amount of energy released when 12 grams of helium is formed.
12/4 = 3 mole of helium = 3 X 6.023 X 10²³ = 1.807 x 10²⁴ helium nuclei
Total energy given out (binding energy) = 4.122X 10^-12 X 1.807 x 10²⁴ = 7.448 X 10¹² joules

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