# Fusion

## Practice

### practice problem 1

4^{1}_{1}H → ^{4}_{2}He + 2(^{0}_{+1}e + ^{0}_{0}γ + ^{0}_{0}ν)

The mass of the sun is 1.99 × 10^{30} kg, 91% of which is hydrogen. Its power output is 3.85 × 10^{26} W. Determine…

- the mass of four hydrogen atoms
- the mass defect when four hydrogen atoms fuse into one helium atom (in atomic mass units and megaelectronvolts)
- the rate at which the sun's mass is decreasing
- the total mass destroyed if all the sun's hydrogen were converted into helium
- the expected lifetime of the sun (assuming its power output will remain constant)

#### solution

- The mass of four hydrogen atoms is…
4 × 1.007825 u = 4.031300 u

- The mass difference between four hydrogen atoms and one helium atom is…
4 × 1.007825 u

− (4.00260 u) = 0.02870 u (0.02870 u)(931 MeV/u) = 26.72 MeV

- The rate at which mass is destroyed is related to the rate at which energy is produced.
*P*=*E*= Δ *mc*^{2}⇒ Δ *m*= *P**t**t**t**c*^{2}Δ *m*= 3.85 × 10 ^{26}W*t*(3.00 × 10 ^{8}m/s)^{2}Δ *m*= 4.29 × 10 ^{9}kg/s*t* - The fraction of the sun's mass lost if every hydrogen nucleus was used to produce helium is the same as the the ratio of the mass defect to the original mass.
*destroyed mass*= *m*= 0.0287 u *initial mass*(0.91)(1.99 × 10 ^{30}kg)4.0313 u *m*= 1.29 × 10^{28}kg - Setting up another proportion finishes the problem.
Δ *m*= 4.29 × 10 ^{9}kg= 1.29 × 10 ^{28}kg*t*1 s t *t*= 3.01 × 10^{18}s = 95 × 10^{9}yearsThis is a not a very good estimate, however. Although extremely hot in human terms, most of the sun is just too cool for hydrogen to fuse into helium. Only the central core is hot enough and dense enough. Thus, it is estimated that not more than ten percent of the sun's total hydrogen will ever be available for thermonuclear fusion. Ten percent of 95 billion years 9.5 billion years, which is still a good long time. The earth is some 4.5 billion years old already, placing us somewhere in the middle of the sun's life. The sun is a middle-aged star.

### practice problem 2

^{3}). These weapons, commonly known as "hydrogen bombs" or "H-bombs", use the energy released when a nuclueus of light lithium and heavy hydrogen, also known as deuterium, fuse to form two nuclei of ordinary helium (a two part process).

^{6}_{3}Li + ^{2}_{1}H → 2(^{4}_{2}He)

A typical thermonuclear weapon has a yield on the order of several million tons of TNT or about as destructive one truck bomb for every person in Brooklyn. (One ton of TNT is equal to 4.184 GJ by definition.)

- Given the reaction described above, determine…
- the mass of one molecule of lithium 6 deuteride
- the mass defect when one molecule of lithium 6 deuteride is transformed into two atoms of helium (in atomic mass units and megaelectronvolts).

- Given a "small" hydrogen bomb with an explosive yield of one megaton, determine…
- the mass destroyed in its detonation
- the mass of the fuel required
- the volume of the fuel required

#### solution

- The mass of one lithium 6 deuteride molecule is simple enough to determine.
(6.015121 u) + (2.0140 u) = 8.0291 u

- The mass difference between the reactants and products of this reaction is also easy to determine.
(6.015121 u + 21.0140 u)

− 2(4.00260 u) = 0.0239 u (0.0239 u)(931 MeV/u) = 22.3 MeV

- Using the mass-energy equivalence gives us the mass destroyed when one of these weapons is detonated. (Recall that the prefix mega means 10
^{6}.)*E*=*mc*^{2}⇒ *m*=*E**c*^{2}*m*=(10 ^{6})(4.184 × 10^{9}J)(299,729,458 m/s) ^{2}

This may not seem like much mass, but it's an enormous amount of energy.*m*= 0.0466 kg - The ratio of the mass of fuel required to the mass destroyed in a one megaton blast is the same as the ratio of the mass of one lithium 6 deuteride molecule to the mass destroyed in a single reaction.
*initial mass*= *m*= 8.0291 u *destroyed mass*0.0466 kg 0.0239 u

This is about half the mass of a medium-size dog.*m*= 15.6 kg - The volume of fuel required can be determined from its mass and density.
*V*=*m*ρ *V*=15.6 kg 820 kg/m ^{3}

This is a bit more than a standard bucket. A bucket of lithium 6 deuteride is sufficient to level all but the largest cities.*V*= 0.0191 m^{3}= 19.1 liters

### practice problem 3

#### solution

Answer it.

### practice problem 4

All these processes, proceeding through microseconds, prepared Mike for thermonuclear burning. Now the escaping X-radiation of the fissioning sparkplug heated the compressed deuterium at its boundaries; the increasing thermal motion of the deuterium nuclei pushed them together until they passed the barrier of electrostatic repulsion between them and came within range of the nuclear strong force, at which point they began to fuse. Some fused to form a helium nucleus — an alpha particle — with the release of a neutron, the alpha and the neutron sharing an energy of 3.27 MeV

^{(1)}. The neutron passed through the electrified mass of fusing deuterons and escaped, but the positively charged alpha dumped its energy into the heating deuterium mass and helped heat it further.Other deuterium nuclei fused to form a tritium nucleus with the release of a proton, the triton and the proton sharing 4.03 MeV

^{(2)}. The positively charged proton dumped more energy into the deuterium mass. The tritium nucleus fused in turn with another deuterium nucleus to form an alpha particle and a high-energy neutron that shared 17.59 MeV^{(3)}. The 14 MeV neutrons from this reaction began to escape the hot, compressed deuterium plasma and encountered the U238 nuclei of the vaporized uranium pusher. U238 fissions when it captures neutrons with energies above 1 MeV; so the U238 of the uranium pusher began to fission then under the intense neutron bombardment, flooding more X rays back into the deuterium mass from the outside just as the sparkplug fission reaction was radiating them from the inside, trapping the deuterium between two violent walls of heat and pressure. Deuterium-bred tritium fused with tritium as well, producing a helium nucleus and two neutrons that shared 11.27 MeV of energy^{(4)}. At lower orders of probability, deuterium captured a neutron and bred tritium^{(5)}; deuterium-bred helium fused with deuterium and made heavy [ordinary] helium plus a highly energetic proton^{(6)}, or captured a neutron and bred tritium plus a proton^{(7)}. All these reactions contributed to the force of the Mike explosion.

#### solution

Seven fusion reactions are described in this passage.

^{(1)} |
2^{2}_{1}H |
→ | ^{3}_{2}He |
+ | ^{1}_{0}n |
||

^{(2)} |
2^{2}_{1}H |
→ | ^{3}_{1}H |
+ | ^{1}_{1}p |
||

^{(3)} |
^{2}_{1}H |
+ | ^{3}_{1}H |
→ | ^{4}_{2}He |
+ | ^{1}_{0}n |

^{(4)} |
2^{3}_{1}H |
→ | ^{4}_{2}He |
+ | 2^{1}_{0}n |
||

^{(5)} |
^{1}_{0}n |
+ | ^{2}_{1}H |
→ | ^{3}_{1}H |
||

^{(6)} |
^{2}_{1}H |
+ | ^{3}_{2}He |
→ | ^{4}_{2}He |
+ | ^{1}_{1}p |

^{(7)} |
^{1}_{0}n |
+ | ^{3}_{2}He |
→ | ^{3}_{1}H |
+ | ^{1}_{1}p |