Homework 1, Physics 3401, Chapter 1, #6,7,12,21,23,31,32,34,36  Rohlf

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 by 3mm.
 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    48.7708 kilograms

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   6.688E+11 kg/s

Go Back