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=  2.8846

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
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 on either side for total of 2.6.

#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. m/s

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

kT= 10^-4 (mc^2/2)  mc^2 =940 Mev
3kT/2= 0.0705 MeV  = 1.13E-20 j
temperature = 5.45 E8 K

#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 =  0.00011        958. meters high
at 250K                    0.00013        798. 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. 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 constant multiplying the exponential is sqrt(m/2 pi kT) =     0.008023
Dist function value = .008023*exp(-3/2)=   0.00179
Multiplying by 2% of the rms speed, 10.3m/s, gives   0.0185
This is the probability for the rms speed - the problem as stated was limited to x direction

For x component only:
vx(rms)=sqrt(kT/m)=298.4 m/s for 28 amu nitrogen
df/dvx = sqrt(m/(2*pi*kT))*exp(-m*vx^2/(2*k*T))
which becomes
=0.001337*exp(-.5)=  0.00081
Multiplying by 2% of the rms speed gives   0.00484

#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 98.8 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=
5.99     1.18E+05     10.2 eV      k = 8.62E-05 eV /K

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

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