6. Density of graphite 2000 kg/m^3 or 2 gm/cm^3
12 grams equals a mole,or 6.02 x 10^23 atoms.
Volume of one carbon atom = 12 grams /(6.02 x 10^23 atoms x 2 grams/cm^3)
9.9668E-24 cm^3/atom 4/3 pi r^3
r^3= 2.3794E-24 cm^3
r= 1.335E-08 cm = 1.33x10^-10 m = 1.33 angstroms
diamter = 0.27 nm compared to Rohlf's nominal 0.3 nm on pg 6
7. Estimated density of copper
Taking the atomic mass number A=64 for copper and the fact that the
atomic mass unit is
1.66 x 10^-27 kg, the mass is estimated to be
m(copper atom) = 1.0624E-25 kg
Taking the nominal diameter to be 0.3 nm from Rohlf p6, the volume 4/3 pi r^3 =
volume = 1.4137E-29 m^3
The projected density is then m/V = 7514.94274 kg/m^3
This 7.5 gm/cm^3 compares to the textbook range of density of copper
= 8.3 - 9 gm/cm^3
12. For orbit radius 100 meters and speed essentially the speed of light.
I = 1 ampere = 1 Coulomb/s=ne/t = nev/(2 pi r)=
1 C/s = n * 1.6E-19C*3E8/(2 pi 100 m)
n= 1.309E+13 electrons
21. Electron of speed 1E6 m/s in field of 1E-4 Tesla.
mv^2/r = evB, so r = mv/eB
r= 0.0569375 meters
23. Electron at 10^6 m/s has same radius as proton; find speed of proton.
m(p)v = m(e)*10^6
10^6/ 1836 = 544.662309 m/s
31. Compare mass energy of mosquito to kinetic energy of 747 jumbo jet.
Approximate mosquito as cylinder of water(or blood) about 0.5mm
Density of water 1 gm/cm^3
This gives mass pi*r^2*L = 5.8905E-07 kg for mosquito
The mass energy is obtained from mc^2 = 5.3014E+10 joules for mosquito
Guessing that a 747 is 50 meters long with a diameter of 10 meters
and a density of 0.2 gm/cm^3
The mass projects to 785398.163 kg
For a cruising speed of 600 mi/hr = 268.224 m/s
The kinetic energy mv^2/2 = 2.8252E+10
So I come out with roughly comparable energies for the mosquito and jumbo jet, from this point of view.
32. One ton of TNT = 4.00E+09 joules according to
Rohlf, from the chemical
energy release of the explosion.
U235 releases 200 MeV of energy from one nucleus.
this is = 3.2E-11 joules
U235 has an atomic mass of 235.04 * 1.66x10^-27 = 3.9017E-25 kg
The required mass of uranium projected from these numbers is 4.8771E-05 kg per ton of TNT
For a megaton, the mass of U235 required would project to
34. Find binding energy of deuteron and U238.
proton mass 0.9383 GeV
neutron mass 0.93957 GeV binding energy of deuterium= 0.00227 GeV
deuteron mass 1.8756 GeV 2.27 MeV
U238 mass 221.697 GeV
mass unit= 1.66E-27 kg
92 protons 86.3236 GeV
146 neutrons 137.17722 GeV
sum of nucleons 223.50082 GeV binding energy of U238 = 1.80382 GeV
which = 7.57907563 MeV/nucleon
36. Fusion energy release = 25 MeV/fusion to yield 4 x 10^26 watts. What mass of hydrogen?
25 MeV= 4E-12 joules
So for 4.00E+26 watts you need 1E+38 fusions/s or 4E38 hydrogens
For proton mass 1.67E-27 kg, this translates to a mass