Physics 3401, Homework Quiz 1

1. Some of the detectors in particle accelerators measure particle energy by the amount of curvature of the charged
particles. If you have 1 MeV protons entering a 1 Tesla magnetic field and being bent into a quarter of a circle, how far along
the beam axis from the entrance port must you put your detector? If you want to use finite detectors such that the difference
in energy at the adjacent detector is 0.05 MeV, how close must the adjacent detector be to the 1 MeV detector?

Using mv^2/r = qvB, the radius of curvature is given by r=mv/qB
Neglecting relativistic corrections, v = sqrt(2E/m) = c*sqrt(2E/mc^2) = c*sqrt(2 MeV/938 MeV) =       13852712.9 m/s
v = 1.39E+07 m/s
Radius of path = mv/qB =  0.145 meters

Doing the same calculation for 1.05 MeV gives v=    1.42E+07 m/s
and radius of path r =   0.148 meters

So detector separation =   0.00365 m = 0.365 cm

2. The element potassium is reported to have one of the larger atomic radii, and may be a counterexample
to Rohlf's universal 0.3 nm atom. If the density of solid is 0.86 gm/cm3 and its molar mass is 39.1 grams,
calculate the atomic radius. State clearly any assumptions you make.

The volume per atom can be calculated from v= molar mass/(density*avogadro's number)
volume = (.0391 kg/mole)/(860 kg/m^3*6.022e23 atoms/mole) =    7.55 E-29 m^3/atom
Taking the volume as a simple cube gives diameter =   4.22648E-10 m = 0.4226 nm

3. If four terrorists make it by airport security on a given day at a sizable airport with 50,000 passengers per day, what
is the probability that you can keep any from getting by that security for a week?

This is a Poisson distribution problem where the probability p for a given day is 4/50000 =      0.00008
For a week the number n is 50,000*7 = 350,000 passengers, so the average expected is np = 28 = a
The probability for zero is exp(-a) = exp(-28) =    6.91 E-13

4. Lay out all the possible ways that 5 particles can have 3 units of energy, and calculate the number of states associated
with each configuration. Find the average population for each state as a way of demonstrating that the population goes down
steeply with increasing energy, leading to the Boltzmann distribution.

 Energy = 0 1 2 3 Weighting Number of part. 4 0 0 1 5 Number of part. 3 1 1 0 20 Number of part. 2 3 0 0 10

Average population
E=0 (4*5 + 3*20 + 2*10 )/35 =  2.857
E=1 (1*20 + 3*10 )/35 =  1.429
E=2 (1*20 )/35 =  0.571
E=3 (1*5  )/35 =  0.143

5. Deuterium-tritium fusion to form an alpha particle and a neutron offers probably the greatest hope for
generating electricity by nuclear fusion. The mass energies of the constituents are given below.

deuterium 1.8756 GeV
tritium 2.8089 GeV
alpha 3.7274 GeV
neutron 0.9396 GeV

Calculate the energy yield of the fusion process and the speed of the lone neutron if it receives all
energy of the process.

Mass difference = 0.0175 GeV= 2.8E-12 joules
v=sqrt(2E/m)=sqrt(2*2.8e-12J/1.6749e-27)=   5.78 E+07 m/s

6. In a helium-neon laser, the excited helium atoms must collide with neon atoms to transfer energy for the "lasing" neon
transition. The gas in the laser tube is 85% helium and 15% neon at about 1/300 of an atmosphere, with helium being
used as a "pumping" gas. If the temperature of the gas is 400 K and 15% of the gas in the tube is neon, how many collisions
per second does a helium atom make with a neon atom? (Although it is not really a valid assumption, you may assume that
both types of atoms have Rohlf's nominal diameter of 0.3 nm. You may also just assume that the gas is helium for the basic
frequency of collision calculation, and then assume 15% of the collisions are with neon.)

Atomic mass of helium = 4.0 u, that of neon = 20.18 u , but that won't be used.

Average speed =  sqrt(8kT/pi*m) = sqrt(8*1.38e-23*400/(Pi()*4.0*1.66e-27))=    1454.98 m/s

In calculating the mean free path, we will have to adjust the volume per mole to account for the pressure of 1/300 atmos.
Volume per mole = 0.0224 m^3 *(400/273)*300 =     9.846  m^3

Mean free path= volume per atom/cross section for collision = 9.846 m^3/(pi*(.3nm)^2*6.02e23*1e-18)

mean free path= 5.78 E-05 m

Rate of collisions = average speed/mean free path =    2.52E+07  /s

Rate of collisions with neon atoms = 15% of collisions =     3.77E+06  /s