practice problem 1
practice problem 2
63Li + 21H → 242He
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)
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 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
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 106.)
E = mc2 ⇒ m = E c2 m = (106)(4.184 × 109 J) (299,729,458 m/s)2
m = 0.0466 kg
This may not seem like much mass, but it's an enormous amount of energy.
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
m = 15.6 kg
This is about the mass of a medium-sized dog.
The volume of fuel required can be determined from its mass and density.
V = m ρ V = 15.6 kg 820 kg/m3 V = 0.0191 m3 = 19.1 liters
This is a bit more than a standard bucket. A bucket of lithium 6 deuteride is sufficient to level all but the largest cities.
practice problem 3
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.
Richard Rhodes, 1995 (paid link)
Seven fusion reactions are described in this passage.