Physics 3401, Homework 5

Rohlf Chapter 5, #1,2,5,7,8,13,17,20,23,24,25,28

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

Stefan's constant=  sigma= 5.67E-08 watt/m^2K^4
Nuclear distance unit  fermi 1.00E-15 m (a femtometer)

#1. Debroglie wavelengths
a. wavelength of photon with energy 10 eV
wavelength =hc/E =  1.24E-07 m
A typical Rohlf approach would be to use hc = 1240 eV nm to
get wavelength =1240 eV nm/10 eV = 124 nm
b. Electron with kinetic energy 6 MeV
This is clearly relativistic, so use wavelength = h/p = hc/pc
Using the relativistic energy expression E=sqrt((pc)^2+(m0c^2)^2)
you get wavelength = hc/sqrt(E^2 -(m0c^2)^2)
where a KE=6MeV gives E=6.511MeV.
wavelength = 1240/sqrt(6.511E6^2 - .511E6^2)=   0.000191036 nm
c. a neutron with momentum 1 keV/c
pc = 1 keV, so wavelength = h/p=hc/pc=1240eVnm/1000 eV = 1.24 nm
d. a neutrino with energy 1 GeV
Since the mass energy is nearly zero if not identically zero, the neutrino
can be treated the same way as the photon in part a.
wavelength =hc/E =1240 eV nm/1E9 eV =    0.00000124 nm
= 1.24 fermi

#2. Kinetic energies for particles of given wavelengths
a. Electron with wavelength 0.1 nm
Try to cast in the units eV and nm.
Kinetic energy = p^2/2m for this non-relativistic electron
KE = h^2/(2m*wavelength)^2 = (hc)^2/(2mc^2*wavelength^2)
=(1240eV nm)^2/(2*0.511e6 eV* (0.1nm)^2)=   150.4500978 eV

If the electron were relativistic we could use E=sqrt((pc)^2+(m0c^2)^2) to
get E and then get kinetic energy from E-m0c^2
This wavelength gives pc=hc/wavelength = 12400 eV
Using rest mass energy .511 MeV we get kinetic energy
150.4279564 eV, so to the accuracy of the constants used, they are the same.

b. photon with wavelength 0.1nm
The photon kinetic energy is given simply by pc=hc/wavelength=1240 eV nm/.1nm
KE= 12.4 keV

c. alpha particle with wavelength 1 fm
The mass of the alpha particle =3730 MeV
For a 1 fm wavelength pc=hc/wavelength=1240eV nm/1E-6nm
pc= 1240000000 eV= 1240 MeV
The energy is then =  3930.712404 MeV
The kinetic energy is then =  200.7124036 MeV
Using the non-relativistic expression p^2/2m , even with relativistic mass and
momentum gives = (1240MeV-fm)^2/(2*3730MeV*(1fm)^2) =    206.1126005 MeV

#5. Electron voltage to be same wavelength as 40 keV x-rays
pc=40keV=hc/wavelength =1.24keV nm/wavelength
wavelength = 0.031 nm
For the electron to have this wavelength, pc=hc/wavelength=1240eV nm/.031 nm
pc = 40000 eV, I.e., pc is  the same for the photon and electron
The electron energy is =  512563.1668 eV
So the electron kinetic energy=  1563.166839 eV

The non-relativistic KE expression p^2/2m gives   1565.55773
so this is essentially a non-relativistic electron.
Bottom line: 1563 volts of acceleration gives electron 0.1nm wavelength

#7. Kinetic energies of electrons, neutrons and photons of wavelength 0.01 nm
Note that the product pc is going to be the same for all three, and that the kinetic energy
can be calculated from pc.
pc= hc/wavelength = 1240 eV nm/0.01 nm = 124000 eV
a. electron
Kinetic energy = E - m0c^2 = sqrt((pc)^2+(m0c^2)^2) -m0c^2
=sqrt(124000^2 + 511000^2) - 511000^2 =   14829.82038 eV

b. neutron
neutron is non-relativistic.
E = p^2/2m = (hc)^2/(2 m0c^2 * wavelength^2) = 1240^2/(2*939.6E6*(.01)^2)
E= 8.182205194 eV

c. photon
For the photon, KE = pc = 124 keV

#8. Debroglie wavelength vs average separation of N2 molecule
a. Taking a temperature of 300K, the average speed of nitrogen molecules
can be obtained from the Maxwell speed distribution
avg speed sqrt(8kT/m*pi)=  476.2883011 m/s

The wavelength can be put in the form wavelength=h/mv=hc/mc^2v
=1240eV nm /(28*931.5E6*476.28)=   9.98185E-11 m

b. The separation between molecules at 300K is about 3.45 nm
so the separation is about   34.56274401 x the wavelength.

#13. Davisson Germer experiment
The Bragg condition for diffraction is n* wavelength = 2d sin theta
The scattering angle is measured to be 50 degrees, so theta = 90-25 = 65 degrees
The electron energy is 54 eV, so we need to calculate wavelength from that.
wavelength = h/p = h/sqrt(2mE) = hc/sqrt(2 mc^2 E)
wavelength = 1240 eV nm/sqrt(2*511000*54) =   0.166916534 nm
Assuming first order, n=1, d = wavelength/(2*sin theta) =    0.092086009 nm

#17. Speed uncertainty of proton located to 1 nm precision.
delta p = hbar/(2 delta x) , delta x = 1 nm
delta v = hbar/(2 m delta x) = c hbar*c/(2 mc^2 delta x)
=3e8 * 197 eV nm/(2* 938E6*1E-9)=   31.50319829 m/s

#20. Electron and proton in atomic-sized volume.
Using the 3D particle in a box expression, the average KE is given by
minimum avg KE = 9 hbar^2/2mL^2
This is for a cube of dimension L, so V=L^3 and L^2=V^(2/3)
For V=10e-30 m^3:    L^2 =  1E-20 m^2
electron avg KE = 5.49269E-18 joules= 34.32929176 eV
proton avg KE= 2.99631E-21 joules= 0.018726937 eV
Note that just the uncertainty principle gives the order of magnitude
of electrons in atoms, whereas it suggests that protons could just wander
in and out of such a volume under the influence of thermal energy.

#23. Fractional width of the tau lepton.
From the uncertainty principle in terms of energy,
lambda = hbar/tau, tau given as 3E-13 seconds
lambda = 6.59E-16/3E-13 =  0.002196667 eV
Since the energy is 1.8E9, the fractional width is 2.2E-3/1.8E9 =    1.22 E-12

#24. Size of confinement correlated with v/c = 0.5.
If you treat the particle like a standing wave in a one dimensional box
so that the wave has value zero at the walls, this gives wavelength = 2L
Then pc=hc/2L and the particle  speed is given by v/c = pc/E=pc/(gamma m0c^2)
For v/c = .5, gamma = 1.15.
The dimension L is than given by L = hc/2pc = hc/2(beta*gamma*m0c^2)
for electron, L=1240MeV fm/(2*.5*1.15*0.511MeV) =    2110.099549 fm
for muon, L=1240MeV fm/(2*.5*1.15*106MeV) =    10.17227235 fm
for charm quark, L=1240MeV fm/(2*.5*1.15*1500MeV) =    0.71884058 fm

#25. Kinetic energy for electron confined to nucleus.
If we take the nuclear size to be 2 fermi
delta p c = hbar c/2 delta x,
delta pc = 197 MeV fm/ 4 =   49.25 MeV
for extreme relativistic case like this , pc approx = energy

#28. Width of the K meson.
From the uncertainty principle in energy,
lambda = hbar/tau , tau = 8.9E-11 sec
lambda=6.59E-16 eV s/8.9E-11 =   7.40339E-06 eV