Question

Deuterium (Subscript 1 Superscript 2 Baseline H) is an attractive fuel for fusion reactions because it is abundant in the waters of the oceans. In the oceans, about 0.0177% of the hydrogen atoms in the water (H2O) are deuterium atoms. (a) How many deuterium atoms are there in one kilogram of water? (b) If each deuterium nucleus produces about 7.20 MeV in a fusion reaction, how many kilograms of ocean water would be needed to supply the energy needs of a large country for a year, with an estimated need of 7.90 × 1020 J?

Answer 1

(a) 1.0 kg of water = 1000 g of water.

No. of moles of water molecules = 1000/18 = 55.6 moles

Now, one mole contains 6.02 x 10^23 atoms/molecules/electrons/ions (i.e. elementary entities)

So, 55.6 moles will contain 55.6 x 6.02 x 10^23 molecules of water in 1 kg of water = 3.34 x 10^25

Therefore, 0.0177 % of the hydrogen atoms are deuterium atoms and the number of hydrogen ATOMS is 2 x 3.34 x 10^25 since there are two hydrogen atoms in each water molecule = 6.68 x 10^25 (3 sig figs)

So, the number of deuterium atoms (in 1 kg of water) is 6.68 x 10^25 x 0.000177 = 1.18 x 10^22 (Answer)

(b) Again we have -

1 eV = 1.6 x 10^-19 J.

So, 7.20 MeV is 7.20 x 10^6 x 1.6 x 10^-19 = 1.15 x 10^-12 J

The energy resulting from the deuterium atoms in one kg of water is = 1.15 x 10^-12 x 1.18 x 10^22 = 1.36 x 10^10 J

Therefore, kilograms of ocean water needed to supply 7.90 x 10^20 J of energy -

= (7.90 x 10^20)/(1.36 x 10^10) = 5.81 x 10^10 kg (Answer)