Preparation problems

44-1) The masses of the neutron, the deuterium nucleus and the He-3 nucleus are 1.008665 u, 2.013553 u, and 3.014932 u, respectively. Use the Eintein mass-energy relation to verify the 3.27 MeV energy release in the D-D fusion reaction of Eq 44-4a.
The reaction in 44-4a is
2H + 2H --> 3He + 1n + 3.27 MeV
1120

Ereleased = Dm c²
= ( mass of reactants - mass of procucts ) c²
= [ (2·2.013553 u) - (1.008665 u + 3.014932 u) ](931.494 MeV/c²) c²
= 0.003269 u · 931.494 MeV = 3.269 MeV

44-36) Estimate the D-D Fusion energy content of Lake Erie, whose volume is 480 km³. How long would this resource supply humanity's energy needs at the current rate of 1013 W?
The density of water is 1000 kg/m³, and in 480 km³ of water there are 4.8 x 1011 m³, so the total mass of Lake Erie is 4.8 x 1014 kg.
The mass of each water molecule is 18 u x (1.66054 x 10-27 kg/u) = 2.99x10-26 kg.
This means the total number of water molecules in Lake Erie is
(4.8 x 1014 kg) / (2.99x10-26 kg/molecule) = 1.6 x 1040 water molecules

Since each water molucule has 2 Hydrogen atoms, that means there are 3.2 x 1040 hydrogen molecules. From the reading, we know that about 0.015% of the Hydrogen nucleii in water are deuterium, so that means we have 4.8x1036 deuterium molecules.
It takes 2 deuterium atoms to make a D-D reaction, so we could have 2.4x1036 reactions.
at 3.27 MeV/reaction, this would release
(3.27 MeV/reaction) x (2.4x1036 reactions) = 1.575 x 1035 MeV = 2.52 x 1024 J.

At the current power usage of 1013 W or 1013 J/s, we could get
(2.52 x 1024 J) / (1013 W or 1013 J/s) = 2.52 x 1011 sec worth of power, or we could power the Earth for about 8000 years!

44-37) The proton-proton cycle consumes four protons while producing about 27 MeV of energy. (a) at what rate must the sun consume protons to produce its power output of about 2 x 1026 W? (b) The present phase of the Sun's life will end when it has consumed about 10% of its original protons. Estimate how long this phase will last, assuming the Sun's 2x1030 kg mass was initially 71% Hydrogen.
2 x 1026 J/sec = 1.25 x 1039 MeV/sec
(1.25 x 1039 MeV/sec) / (27 MeV/reaction) = 4.63 x 1037 reactions/sec
At 4 protons/reaction, this means the sun must use up 1.85 x 1038 protons/sec!

0.71 x (2 x 1030 kg) = 1.42 x 1030 kg of protons in the sun initially.
Each proton has a mass of 1.66 x 10-27 kg, so there were (1.42 x 1030 kg) / (1.66 x 10-27 kg/proton) = 8.55 x 1056 protons.
At the rate of 1.85 x 1038 protons/sec these will be used up in
(8.55 x 1056 protons) / (1.85 x 1038 protons/sec) = 4.6 x 1018 sec = 146 Billion Years!

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