# Nuclear fusion

this is homework problem i'm having some trouble with... i haven't had chemistry yet so i'm not sure how i could convert atoms into kg, and the textbook i have doesn't do any examples of problems like this one... i don't even know where to begin.

Deuterium (A=2, Z=1) is an attractive fuel for fusion reactions because it is abundant in the waters of the oceans. In the oceans, about 0.015% of the hydrogen atoms in the water (H2O) are deuterium atoms.

(a) How many deuterium atoms are there in 4 kilogram of water?

(b) If each deuterium nucleus produces about 7.4 MeV in a fusion reaction, how many kilograms of ocean water would be needed to supply the energy needs of the United States for one year, estimated to be 9.3 1019 J? (answer should be in kg)

HallsofIvy
Homework Helper
mayo2kett said:
this is homework problem i'm having some trouble with... i haven't had chemistry yet so i'm not sure how i could convert atoms into kg, and the textbook i have doesn't do any examples of problems like this one... i don't even know where to begin.

Deuterium (A=2, Z=1) is an attractive fuel for fusion reactions because it is abundant in the waters of the oceans. In the oceans, about 0.015% of the hydrogen atoms in the water (H2O) are deuterium atoms.

(a) How many deuterium atoms are there in 4 kilogram of water?
How many hydrogen atoms are there in a kilogram of water? You should be able to get that by using the molecular weight (in kg/molecule) of water: the atomic weight of oxygen plus two times the atomic weight of hydrogen. Be sure to use kg/molecule! Multiply by 2 to get the number of hydrogen atoms in one kg of water (there are 2 hydogen atoms per water molecule of course). Multiply by 4 to get the number of hydrogen atoms in 4 kg of water. Finally, multiply by 0.015%= 0.00015 to get the number of deuterium atoms in 4 kg of water.,

(b) If each deuterium nucleus produces about 7.4 MeV in a fusion reaction, how many kilograms of ocean water would be needed to supply the energy needs of the United States for one year, estimated to be 9.3 1019 J? (answer should be in kg)
Divide your previous answer by 4 to get the number of deuterium atoms in one kg of water and multiply by 7.4 MeV to get the energy, in MeV, in one kg of water. Now you will have to convert 9.2x1019 J to MeV to get the energy required in MeV. Finally, of course, divide the energy in MeV by the energy per kg of water to get your answer.

a little confused

i tried the first part but i'm not sure if this is what you meant...

molecular weight of water = 18 g/mol = .018 kg/mol

then:
2(.018kg/mol) = .036 hydrogen atoms in 1 kg of water

4kg(.036) = .144 hydrogen atoms in 4 kg of water

.00015(.144) = 2.16e-15 number of deutrerium atoms in 4kg of water

don't i need to use the avagadros number someplace?

Astronuc
Staff Emeritus