2.1 a) Number of atoms in a Yoctomole = 1e-24 * (6.022e23) ≈ .6 atoms

2.1 b) Number of seconds in a nanocentury = 1e-9 * 100 * 365 * 24 * 60 * 60 = 3.15 ≈ π

2.2) Number of CDs for a petabyte is 1e15 bytes * (700e6 bytes)^{-1} ≈ 1.43e6. The thickness of a CD is 1.2 mm, so the height is 1.71 km, about twice as high as the tallest building in the world.

2.3) 1e80 atoms in the universe, so the number would be about 2^{80} ≈ 1.21e24.

2.4) Gravitational acceleration on earth's surface ≈ 9.8 m/s^{2}. Gravitational acceleration due to a 1 kg mass 1 meter away is: G * (1 kg) / (1 m^{2}) ≈ 6.6e-11 m/s^{2}. The ratio is: 223.3 dB.

2.5 a) Looking up the heat of formation of TNT, it is -54.39 kJ/mol. After it combusts, it produces gaseous components with a heat of formation of 6*-111.8 = -670 kJ/mol. The net result is 616.4 kJ/mol, so we would expect one ton of TNT to produce 2000 lbs / (2.2 kg/lb) / (227.13 g/mol) * (616.4 kJ/mol) = 5e9 Joules.

2.5 b) Looking at the mass that is converted to energy in a uranium explosion, it is 3.3e-25 g. This equivalent to 2e-11 Joules per atom of Uranium. For 10,000 tons, you would need just 10000 * 2000 lbs / (2.2 kg/lb) / (227.13 g/mol) * (616.4 kJ/mol) / (0.2 amu * speed of light * speed of light) / 6.022e23 = 0.28 mols only!

2.5 c) That is 1/120th of the amount of total energy in 0.28 mols of Uranium.

2.6 a) The de Broglie wavelength is h/mv ≈ 6.63e-34 Joule*seconds / (.2 kg * 10 m/s) ≈ 3.32e-34 meters.

2.6 b) Using kT = mv^{2}, we have mv = (kTm)^{.5} ≈ (1.38e-23 J/K * 300K * 2.3 e-23 g)^{.5} ≈ 9.76e-24 m*kg/s. So the wavelength is ≈ 6.78e-11 meters.

2.6 c) Using PV = nRT, at room temperature, 1 mol of gas will take up about 24.6 L. To compute the average distance, we can divide (.0246 m^3/6.022e23)^{1/3} ≈ 3.45e-9 meters.

2.6 d) We would like to have (RT/P/6.022e23)^{1/3}=h/(kTm)^{1/2}. Solving that for T, we find that T≈ 2.7 Kelvin.

2.7 a) If potential energy is -GMm/r, then kinetic energy required is 1/2mv^{2}, so escape velocity is (2GM/r)^{1/2}

2.7 b) Solving for r when v is the speed of light, we find that the radius is 2GM/(c^{2}) = M/(6.73e26 kg/m)

2.7 c) If a mass M is converted to a photon, it's energy is 1/2Mc^{2} = hf = hc/l (where f is frequency and l is lambda). So l = hc/(1/2Mc^{2}) = 2h/Mc.

2.7 d) Setting 2h/Mc = M/(6.73e26 kg/m), we see that M = 5.45e-8 kg.

2.7 e) The size is very small, 2h/Mc = 8.1e-35 meters.

2.7 f) Yet the energy is 1/2Mc^{2} = 2.45e9 Joules.

2.8 a) We can take a cross-section of the pyramid and turn this problem into a circle inscribed inside an isosceles triangle.