Homework 1, Physics 3401, Chapter 1, #14,18,22,23,31,32,34,36  Rohlf

Physical constants

Avogadro's number  A 6.02E+23
Boltzmann's constant  k= 1.38066E-23 J/K
  k= 0.00008617 eV/K
Atomic mass unit   u= 1.66E-27 kg
  umev= 931.5 MeV/c^2

Problem 1: Calculate atom size for gold
 density of gold  19300 kg/m^3
 atomic mass of gold  197

 atomic volume = molar mass/(density*Avogadro's number)
 atomic volume = (.197 kg)/(19300 kg/m^3 * 6.02x10^23 atoms/mole)
 atomic volume = 1.69556E-29 m^3/atom
 Taking the cube root of this volume gives d =   2.57E-10 m= 0.257 nm
 Treating volume as a sphere gives d=   3.19E-10 m= 0.319 nm

 Twice covalent radius from periodic table:   0.268 nm

Problem 2:Calculate density of aluminum if its diameter is the nominal 0.3nm .
 atomic mass of aluminum = 27
 density = molar mass/(atomic volume*Avogadro's number)
 density =   (27/((0.3e-9m)^3*6.02e23)
 density =   1666.67 kg/m^3
 Actual density = 2700 kg/m^3   density =   3423.548659 kg/m^3
    using covalent diameter 0.236nm from periodic table.
Problems from Rohlf:
14. A. Kinetic energy of hydrogen orbit, assuming circular.
Potential energy -ke^2/r
Force relationship, centripetal acceleration:  mv^2/r = ke^2/r^2
Kinetic energy mv^2/2 =(r/2)*(mv^2/r)= ke^2/(2r) = -PE/2

b. Size of orbit
  -13.6eV = -1.44 eV nm/(2*r)
r=  0.0529 nm

c. Speed of electron in orbit
KE=mv^2/2=13.6 eV
v=sqrt(2*13.6eV*1.6e-19J/eV/(9.11e-31kg))=   2.19E+06 m/s= 0.00729 c

18. A. Gravity negligible because it is down to about 10^-39 x electric force.
b. The electron velocity in Ex 1-4 is given to be 2.9e7 m/s, and the distance specified
is L=0.050 meters.
The time of transit is then L/v = 0.05/2.9e7=   1.72414E-09 sec
The gravitational deflection is then 0.5gt^2 =0.5*9.8*(1.72e-9)^2=    1.44962E-17 m
So the gravitational deflection is quite negligible, being on the order of 1/100 the nuclear size.

22. Radius of curvature of 10keV electrons in 1 milliTesla field.
r= mv/qB for charge in magnetic field.
v=sqrt(2E/m)=sqrt(2*1e4 eV*1.6e-19J/eV/9.11e-31kg)=    5.9267E+07 m/s
r= (9.11e-31 kg*5.927e7m/s/(1.6e-19C*1e-3T)=    0.337 m

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 = 544.66 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 =   5.89049E-07 kg for mosquito
 The mass energy is obtained from mc^2 =   53014376029 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  785398.1634 kg

 For a cruising speed of 600 mi/hr =   268.224 m/s

 The kinetic energy mv^2/2 =  28252387571
 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 =    3.90166E-25 kg

The required mass of uranium projected from these numbers is     4.87708E-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