Homework 2, Physics 3401, Chapter 2, #7,8,9,10,13,15,19,23,24,27,29,32,34  Rohlf

#7 Full width at half maximum for Gaussian distribution

 The form of a Gaussian distribution about x=0 is

 f = C exp(-x^2/(2 sigma^2))

 If it is half maximum at x=h, then

 0.5C = C exp(-h^2/(2 sigma^2))

 ln(.5)=-h^2/(2 sigma^2)


 h = sqrt(2ln2)sigma

 full width at half maximum = 2h =2.355 sigma

#8 Calculate the probability of x Red Sox cards from 75 cards
distributed randomly among 26 teams.

 Use Poisson distribution  f(x)=exp(-a)*a^x/x!

 n=75, p=1/26, a= 75/26=  2.884615385

Poisson Binomial 
For x= f(x)= f(x)= 
0.055874  0.052781
1 0.016117  0.015834 
2 0.23247   0.20802
3 0.22353  0.22811
0.1612 0.16424
5 0.093 0.09329
6  0.044713 0.043536

 The standard deviation is 1.66

#9 CERN Z boson detection
 If 5 decays are observed in n events,
 what is prob that at least 2 more will be observed in another n events.

 The statistics of small numbers suggests the Poisson distribution.

 Since we have nothing else to go on, assume that 5 is normal, or mean for n

 Find probability of 0 and 1 and subtract from 100%

 f(x)=exp(-a)*a^x/x!  with x = 0 and x=1

 x f(x)
 0 0.006738
 1 0.033689

 sum 0.040427
 So probability of at least 2 is .96 or 96%

#10 Probability of exactly 500 heads out of 1000 coin tosses.
 The large number suggests Gaussian distribution

 Binomial 0.025225
 Gaussian 0.025231
 Poisson 0.017838

 So with just  1000 coin tosses, the Gaussian
 distribution is good to almost 4 significant digits,
 we can be fairly confident in it when we have
 Avogadro's number of molecules.

#13. One event in 10^6. How many to be 90% sure of a second?

 Small probability, big numbers of events.
 Tailormade for the Poisson distribution.

 Probability of seeing zero events is exp(-a) = 0.1 or 10%

 This gives a= 2.3
 If p is equal to 1 in 10^6, then we need 2.3 times that to give a=2.3
 So 2.3 x 10^6 more events to be 90% sure.

#15. Elastic scattering of 10 GeV electrons into a detector with standard deviation of 10%.

 a. Bell curve of normal distribution centered on 10 GeV
 n=1000, sigma=1 GeV, a=10
 Peak of curve is at f(x) = 1/sqrt(2 pi) = 0.4, or 400 on vertical axis out of 1000

 b. Check normal curve for percentage outside 3 standard deviations.
 The fraction from the curve is .0013, or 1.3 events out of 1000.

#19. Estimated collision rate for nitrogen molecules at STP

 Take average speed divided by the mean free path
 If mass of 28 amu and diameter of 0.3 nm is used,
 the collision rate is  4.88 x 10^9/s
 For diameter = .25nm, the rate is   3.39 x 10^9/s

 For the 0.3 nm nominal diameter
 average speed 454 m/s
 mean free path 9.3 x 10^-8 meters

#23. a. Typical speed of a thermal neutron
 kT= 1/40 eV = mv^2/2 for neutron
 3kT/2=.04 eV = 0.621 x 10^-20 joules 6.21E-21
 m =  1.68E-27
 v=sqrt(2E/m)=  2723.036058 m/s

     b. At what temperature is speed equal to 1% of speed of light

 kT= 10^-4 (mc^2/2)  mc^2 =.511 Mev
 3kT/2= 38.325 eV  = 6.14043E-18 j
 temperature = 300000 K 2.96E+05

#24. Protons on sun at 6000K. Find average KE and most probable speed

 kT= mv^2/2 gives average KE   3kT/2 = 0.776 eV
 Boltzmann dist gives most probable speed
 For mass of 1 amu  Most probable speed from Boltzmann =   10000 m/s
#27. Change in altitude to get 10% pressure reduction in nitrogen at 300K,250K

 Barometric formula  0.9 = exp(mgh/kT)   -mgh/kT =ln(.9) =
 28 amu for nitrogen

 u 1.66E-27 kg
 g 9.8 m/s^2
 k 1.38E-23 J/K
 at 300K, mg/kT =  0.000109973  958.0621655 meters high
 at 250K  0.000131967  798.3851379 meters high

#29. Fraction of molecules within 1% of the rms velocity, gas of nitrogen molecules.

 Given room temperature  300 K
 nitrogen mass  28 amu
 k  8.62E-05 eV/K
 vrms = sqrt(3kT/m)  517.0405504 m/s
 u  1.66E-27 kg
 e  1.60E-19 J/eV
 Plugging in the expression for vrms into the Maxwell speed distribution
 the exponential becomes exp(-3)
 The probability is the distribution functioin x 2% of the rms speed
 The constant multiplying the exponential is sqrt(m/2 pi kT) =     0.001336408
 The probability is then =  0.000688035

#32. Vacuum tube filament with work function 2 eV. Electron flux at 1000K.

 C(A/(m^2eV^2) 1.60E+14
 k 8.62E-05 eV/K
 T 1000 K
 kT 8.62E-02 eV
 (kT)^2 0.007425269
 C(kT)^2 1.19E+12
 work fcn/kT 2.32E+01
 Current dens 9.88E+01 A/m^2
 e 1.60E-19 coulombs
 electron flux 6.18E+20 electrons per sec per square meter

#34. At what temperature would 1% of hydrogen electrons be in 1st excited state.

 relative population
 ln(400) delta E/k energy diff in ev k=
 5.991464547 1.18E+05 10.2 8.62E-05 eV /K
ln(400) delta E/k energy diff in ev k=
5.991464547 1.18E+05 10.2 8.62E-05 eV /K

 T=(DE/k)/(4*ln(10))  1.98E+04 K = temperature