Physics 3401, Homework 7
Due Monday, November 16
Rohlf Chapter 7, #1,6,8,13,15,17,24,26,28,31,33

Physical constants
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
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
 h/ 2 pi hbev 6.59E-16 eV s
 hc 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  mel 9.11E-31 kg
  mel 5.11E-01 MeV
proton mass  mp 1.67E-27 kg
  mpev 9.38E+02 MeV/c^2
neutron mass  mn 1.67E-27 kg
  mnev 9.40E+02 MeV/c^2
Stefan's constant=  sigma= 5.67E-08 watt/m^2K^4
Nuclear distance unit  fermi 1.00E-15 m (a femtometer)
cross section unit  barn= 1.00E-28 m^2
  barn= 100 fm^2
common combo for scattering  ke^2 1.44 MeV fm
  hbar*c 3.16138E-26 J m
  hbar*c 197.5863737 eV nm
  h*c 1.98635E-25 J m
  h*c 1241.4718 eV nm
#1  Wavefunction for a 100 eV free electron
wavefunction = A*exp (i(px - Et)/hbar)
E = 100 eV, p=sqrt(2 m E) , approx 10 keV/c
p= 10109.40156 eV/c

#6. Proton in 2 fermi one dimensional box, first two energies
E = (nh)^2/(8mL^2) = (nhc)^2/(8 mc^2L^2) =   5.12E+01 MeV for n=1
   2.05E+02 MeV for n=2

#8. Electron in box, 10 eV difference between ground and first excited, find L
Since the energy is proportional to n^2, the difference between the ground and first excited
states gives  3(hc)^2/(8 mc^2L^2) = 10 eV
L = sqrt(3)*hc/sqrt(80*mc^2)  0.335913044 nm

#13. Finite square well of height 10 eV. If ground energy is 1 eV, what is width of well?
The equation for the finite square well is trancendental if you are solving for energy, but
in this case the energy is given, so you can solve for the corresponding well width L.
tan(sqrt(2*m*E)*L/(2*hbar)) = tan(sqrt(2*mc^2*E)*L/(2*hbar*c))=sqrt((V0-E)/E)
tan(sqrt(2*.511E6*1)*L/(2*197)) = sqrt((10-1)/1) = 3
sqrt(2*.511E6*1)*L/(2*197) =  1.249045772
L = 0.486798384 nm

#15. Alpha particle of energy 4 MeV, Estimate lifetime.
From Fig 7-17, p208 of Rohlf, read halflife of about 1E18 seconds

#17. Probability for 3 eV electron penetrating 4 eV barrier of height 4 eV and width 0.1 nm.
The probability for penetration is of t he form exp(-2 beta x) where x is the thickness of the barrier.
The term beta is given by beta=sqrt(2 mc^2 (V0 - E))/hbar*c
beta = 5.131675921 nm^-1
The corresponding lifetime is exp(-2*5.13 nm^-1*0.1nm) =    0.358317726 seconds.

#24. Uncertainty principle and the quantum harmonic oscillator
a. delta x = sqrt(hbar/ m*omega)
b. delta p = sqrt(2*m*E) = sqrt(2*m*(hbar*omega/2)) = sqrt(m*hbar*omega)
c. For an electron where hbar*omega = 10 eV
delta x = sqrt(hbar^2*c^2/(mc^2 hbar*omega))= hbar*c/sqrt(.511E6 eV*10 eV)
delta x =  0.087147669 nm
delta p= sqrt(m*hbar*omega)=sqrt(mc^2* hbar*omega)/c =sqrt(.511E6 eV*10 eV)/c
delta p=    2260.530911 eV/c

#26. Energy for electron in atomic sized 3-D box.
Let the wavelength of the electron equal twice the dimension of the box so that
wavelength = 2*V^(1/3) = 2*(1E-29)^(1/3)=   4.30887E-10 m
minimum momentum = sqrt(3(h/(2*V^(1/3))^2) = sqrt(3)*hc/(2*c*V^(1/3))
 =sqrt(3)*1240 eV nm/(2*c*(10^-2 nm^2)^(1/3)) =    4984.469966  eV/c
b. The ground state energy is E=3*(hc)^2/(8*mc^2*V^(2/3)) =
 =3*(1240 eV nm)^2/(8*.511E6 eV*1E-2 nm^2)^(2/3) =    24.31011824 eV

#28. Electron with energy 13.6eV in 3-D box.
L=sqrt(3*(1240 eV nm)^2/(8*.511E6*13.6 eV)) =    0.288043059 nm

#31. Potential well, calculate number of bound states.
For a potential well problem, the energies are just the particle in a box energies minus the depth of the well.
En = - V0 + (nh)^2/(8m(2L)^2)
The coefficient of n^2 is (hc)^2/(8 mc^2 (2L)^2) = (1240 eV nm)^2/(8*.511E6 eV*(.2 nm)^2)
9.403131115 eV
The n=1 state is then -10 + 9.4 eV = -0.6 eV, and the n=2 state is positive,
so there is only one bound state.

#33. Estimate ground state energy of electron in nucleus.
Approximating the nucleus as  an infinite square well of dimension 5 fermis,
then we can take the wavelength to be on the order of 5 fermis
The momentum is then p=h/wavelength and pc=hc/wavelength
This gives pc= 1240MeV fm/5 fm =   248 MeV
Since this is far above the rest mass energy of 0.511MeV, we see that the energy is approx 248 MeV.

Try to develop a routine for solving the finite well.

For an infinite well, the energies are given by   E(n)=(nh)^2/(8mL^2)

 n h lm lnm m mev
 1 6.63E-34 2.00E-10 0.2 9.11E-31 5.11E+05

 Infinite well energy =  1.50585E-18 joules = 9.403131115 eV

 For the finite well, even solution:
 For well depth  U0 =
  1.00E+03 eV 1/beta Effective width Energy for effective width
 First approx 9.403131115  0.006209878 0.212419756 8.335711438
  8.335711438  0.006206535 0.21241307 8.336236208
  8.336236208  0.006206537 0.212413073 8.33623595
  8.33623595  0.006206537 0.212413073 8.33623595
  8.33623595  0.006206537 0.212413073 8.33623595
  8.33623595  0.006206537 0.212413073 8.33623595
  8.33623595  0.006206537 0.212413073 8.33623595

 Alpha decay of Uranium-238
 The alpha from U-238 decay has an energy of 4.27 MeV and a half-life of 4.46 billion years.
 The height of the barrier is calculated to be about 35 MeV
 The radius of the U-238 nucleus is about 7.4 fermi.
 The value of of the potential drops to the alpha energy at distance
  r = 7.4 fm *(35/4.27) =  60.6557377 fm
 Try to calculate a decay constant based on 5 segments with barrier height equal to the midpoint potential.
 Width of segment =  10.65114754
 x V(x)-E beta exp(-2*beta*x) net
 5.32557377 16.08271687 1.752920345 7.78918E-09
 15.97667377 6.80942056 1.140610991 5.29493E-06 4.12431E-14
 26.62777377 3.341429468 0.799003391 0.00020139 8.30597E-18
 37.27887377 1.526923202 0.540121009 0.003173618 2.636E-20
 47.92997377 0.411007099 0.280225166 0.050554351 1.33261E-21

 With 9.7E20 hits of alpha at radius, this gives a decay constant =    1.72258E-21
 And a half-life of  4.02304E+20 sec= 1.27482E+13 years