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
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
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
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
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
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