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^2=-2ln(.5)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     0.055874    0.052781
1     0.16125      0.15842
2     0.23247      0.20802
3     0.22353      0.22811
4     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

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

Small probability, big numbers of events.

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)=  2723.0 m/s

b. At what temperature is speed equal to 1% of speed of light    v^2 rms = 3kT/m, so if v=0.01c, then 3kT/m=.0001c^2
or 3kT/2 = 0.0001*mc^2/2
kT= (10^-4)*((mc^2)/3)  mc^2 =940 Mev
3kT/2= 0.047 MeV  = 7.53034E-15 j
temperature = 3.63E+08 K
If you use kT instead of 3/2 kT you get T =   5.45E+08

#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
9989
#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) =  -0.105360516
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.

Find the value of the function f=dn/df, the derivative of the probability with respect to frequency.
Then the probability is then approximated by dn = fdf where f is evaluated at v(rms) and df=0.02v(rms)

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(-3/2)
The probability is the distribution functioin x 2% of the rms speed
The term multiplying the exponential is 4pi*(m/2 pi kT)^3/2*v^2 =     0.00802
The probability is then =  0.018514024

For x component only:
vx(rms)=sqrt(kT/m)=298.4m/s for 28 amu nitrogen    298.44 m/s
df/dvx = sqrt(m/(2*pi*kT)*exp<-m*vx^2/(2*k*T))    0.00335
which becomes
= 0.00335068*exp(.5)  0.005524337
Multiplying by 2% of the rms speed gives   0.0329

#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
98.83 amps/m^2

#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

T=(DE/k)/(4*ln(10))  1.98E+04 K = temperature
2.57E+04 if you leave out the 8/2 relative weighting.

Energy distribution problem: 5 particles with 4 units of energy
Weight
Energy 0 1 2 3 4
4 0 0 0 1 5
3 1 0 1 0 20
3 0 2 0 0 10
2 2 1 0 0 30
1 4 0 0 0 5

Average population of energy states
0 2.5
1 1.43
2 0.71
3 0.29
4 0.07