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(3h^2/(8*m*E))=sqrt(3(hc)^2/(8*mc^2*E))
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