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

your criteria for your answer.

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