Physics 3401, Quiz 3,  Chapters 5,6 Rohlf

1. An electron, a photon, and a proton are each given an energy of 2 GeV (kinetic energy for the particles.)
Find the deBroglie wavelength associated with each. Which of the particles is most relativistic? Show

w=hc/pc= 1240ev nm/mc
For electron, pc= sqrt((2000+.555)^2-.511^2) =   2000.55 MeV, so you could have just used pc=E
w = 1240 MeV fm/2000.55 Mev=  0.6198 fm

For proton, pc=sqrt((2000+938)^2-938^2)=   2784.2 MeV
w = 1240 MeV fm/2784.24 Mev=  0.4453 fm

For photon, pc=E, so w = 1240MeV fm/2000 MeV=   0.62 fm

2. In order to probe the structure of the atom, it is reasonable to  have a wavelength about a tenth of the
size of the atom, say 0.03 nm.
Find the necessary energies for protons, electrons and photons to achieve this wavelength.

For photon, pc=E, pc = 1240 eV nm/.03 nm =   41333. eV

pc is the same for the other particles, so just the energy must be calculated.
For the proton,  KE = sqrt((.041333)^2+938^2)-938 =    9.10E-07 MeV
=  0.91 eV
For the electron,  KE = sqrt((.041333)^2+.511^2)-.511 =    0.001668 MeV
1669 eV

3. It turns out that electrons are the most versatile probe of nuclear structure since they are attracted
rather than repelled, and  they do not break down into anything smaller.

a. Suppose you want to probe to one tenth the size of a hydrogen nucleus, radius 1.2 fm.  What energy
electrons would be necessary?

For electron, pc = 1240 MeV fm/.12 fm =   10333. MeV
For the electron,  KE = sqrt((10333.333)^2+.511^2)-.511 =    10332.8 MeV

b. Suppose at a much larger scale, you want to examine the lattice geometry of a crystal where the
separation between atomic planes is 1.2 nm. If you want the electron to have a deBroglie wavelength twice
the lattice spacing, what energy electrons would be necessary?

For electron, pc = 1240 eV nm/2.4 nm =   516.6 eV
For the electron,  KE = sqrt((516.66)^2+511000^2)-511000 =    0.261 eV

4. A color television electron gun gives the electrons about 23000 eV of energy. The electrons have to go
through apertures of diameter 0.5mm in the metal picture tube mask in order to hit the correct phosphor
dot. If the uncertainty in energy is 100 eV, what is the corresponding uncertainty in position given by
the uncertainty principle?

pc=sqrt((23050+511000)^2-511000^2)=   155204. eV
pc=sqrt((22950+511000)^2-511000^2)=   154859. eV
delta pc =   344.4 eV
delta x = hbar c/2 delta pc = 1240 eVnm/4*PI()*344 eV =    0.286 nm

So the position uncertainty for the electrons is down at atomic dimensions, so you don't have to
worry about quantum properties of the electron for the color tv application.

5. If the uncertainty in momentum Dp is taken  to be equal to the momentum p, a common presumption
for central force problems, then, using the uncertainty principle in one dimension

a. What energy would be required to contain an electron inside an atom of size 0.3 nm (i.e., Dx=0.3nm)
delta pc=  hbar c/2 delta x = 197 eV nm/(2*0.3 nm)   328.33 eV
E = (pc)^2/2mc^2 = (328.333)^2/(2*511000eV) =    0.10548 eV

For proton in nucleus
delta pc = 197MeV fm/2*0.11.4fm =   8.640 MeV
E=(pc)^2/2mc^2 =   0.0398 MeV

For electron in nucleus
delta pc = 197MeV fm/2*0.11.4fm =   8.640 MeV
E=sqrt((pc)^2+(mc^2)^2) =   8.655 MeV

6. A radioactive source sends a million alphas per second toward a 1x10-6 m gold foil. A graduate student
watches a 4 cm2 area of scintillator screen for flashes indicating the impace of an alpha. The screen is
centered at angle 30° at a distance of 20 cm from the foil so that it covers angles from 27° to 33°. Predict
the number of flashes per second the graduate student will see. For gold, Z=79, density 1.9x104 kg/m3, A= 197.

Cross section for above 27° =   233.18 barns
Cross section for above 33° =   153.17 barns
Difference  80.01 barns
Scattered fraction in angular range 27 to 33:
Scattered fraction = (6.02e23)*(1e-6)*19000*80e-28/(197e-3) =     0.000464

Fraction of circle captured = 2cm/2*Pi()*20*sin(30) =    0.03183

So out of a  million per second, you should see     14.78  /s

7. With a detector at 30° as in the problem above, calculate the closest approach to the nucleus for a
5.5 MeV alpha like the ones Geiger and Marsden presumably used.
The impact parameter is given by
b= ((1.44 Mev fm)*2*79/2*5.5MeV)*sqrt((1+cos(30*Pi()/180))/(1-cos(30*Pi()/180))) =
b= 77.19238179 fm
The closest approach is then rmin=b*cos(15*Pi()/180)/(1-sin(15*Pi()/180))
rmin = 100.6 fm

b. If the gold nuclear radius is 7 fm, what kinetic energy would be required to just graze the nucleus?

rmin = 7fm would correspond to an impact parameter
b = (7fm)*(1-sin(15*Pi()/180))/cos(15*Pi()/180) =    5.371 fm

This would correspond to a kinetic energy
KE = ((1.44 Mev fm)*2*79/(2*b))*sqrt((1+cos(30*Pi()/180))/(1-cos(30*Pi()/180)))
KE= 79.04 MeV