Solution for Chapter 1 Quiz, F99

Physical constants
Boltzmann's constant  k= 1.38066E-23 J/K
                                    k= 0.00008617 eV/K
Atomic mass unit   u= 1.66E-27 kg
                                 = 931.5 MeV/c^2
Wien constant   2.90E-03
Planck's constant  h= 4.14E-15 eV s
                              h= 6.63E-34 J s
 h/ 2 pi = hbar =1.05E-34 J s
                        = 6.59E-16 eV s
hc =1.24E+03 eV nm
Speed of light  c= 3.00E+08 m/s
Electron charge  e= 1.60E-19 C
electron volt  eV= 1.60E-19 joule
electron mass   = 9.11E-31 kg
                         =  5.11E-01 MeV
proton mass      = 1.67E-27 kg
                         = 9.38E+02 MeV
Stefan's constant=  sigma= 5.67E-08 watt/m^2K^4
Nuclear distance unit  fermi = 1.00E-15 m (a femtometer)

1. Many of the more massive stars get their energy from a fusion cycle called the "carbon cycle". Assume
 that the path to carbon just involves the fusion of 12 protons to form carbon-12 (which has a mass of
 exactly 12 amu). Never mind how it gets rid of the 6 units of charge - just calculate the energy
 yield from converting 12 protons to a carbon-12 nucleus.

Mass of 12 protons = 938 * 12                       =    11256 MeV
Mass of carbon-12 nucleus = 12*931.5 MeV =    11178 MeV
               Energy yield per carbon                                78 MeV

2. The hottest experiment in high energy physics right now is the planned collisions of gold nuclei at
Brookhaven in a ring of circumference 2.4 miles. Taking the mass of the gold nucleus to be
197 amu and assuming it to be stripped of electrons (i.e., positive charge +79), what magnetic field
 will be necessary to accelerate the gold nuclei in this 615 meter radius circular path. Assume the
velocity to be the speed of light - it's close enough.

F=mv^2/r = qvB,   so B = mv/qr   m=197*1.66E-27kg/amu =  3.2702E-25 kg

B= (3.27E-25 kg)(3E8 m/s)/(79*1.6E-19 C*615 m) =     1.26E-02 Tesla
                                                                                            = 126 Gauss

3. Taking Rohlf literally in his approximation that all atoms have diameters of 0.3 nm, what density do you
get for aluminum which has an atomic mass of 27? I've alway thought this 2.7 gm/cm3 = 2700 kg/m3
was a bit strange.

Approximate by cubic volume = cube of 0.3 nm =     2.7E-29 m^3
Mass = 27*1.66E-27 Kg =    4.482E-26 kg
Density   =                             m/V = 1660 k/m^3