Physics 3401, Exam 1

Physical constants
Boltzmann's constant  k= 1.38066E-23 J/K
k= 0.00008617 eV/K
Atomic mass unit   u= 1.66E-27 kg
Wien constant   2.90E-03
Planck's constant  h= 4.14E-15 eV s
h= 6.63E-34 J s
Speed of light  c= 3.00E+08 m/s
Electron charge  e= 1.60E-19 C
electron volt  eV= 1.60E-19 joule
electron mass  mel 9.11E-31 kg
mel 5.11E-01 MeV
Stefan's constant=  sigma= 5.67E-08 watt/m^2K^4

1. a. Probability of 4 heads out of 10 coin flips
This is a case which requires the binomial distribution because of
small number and high probability for each event.

p=.5, n=10, and x=4
The probability is then =  0.205

b. Probability of 400 heads out of a 1000 coin flips.
Because of the large number, this requires the gaussian distribution
Probability =  5.20E-11

2. Rms speed can be obtained from setting 3kT/2 = average kinetic energy, the kinetic
temperature definition.
3kT/2 = 6.21297E-21 J at 300K
at 29amu, the average molecule mass is    4.814E-26 kg
the speed squared is then =   258120.8974 m^2/s^2
the rms speed is then =  508.0559983 m/s

3. a. Photon energy at peak of 6000K blackbody curve
a. Using Wien's displacement law, the peak wavelength is     4.83E-07 m
The photon energy =hc/wavelength =    4.11E-19 joules
2.57E+00 eV
b. For 10% of the power emitted by one square centimeter
Power = 0.1*sigma*1E-4 m^2*T^4 =   7.35E+02 watts

4. Relativistic kinetic energy and momentum of electron at 0.9c
gamma is equal to   2.294157339
Kinetic energy =  0.6613144 MeV = 1.0581E-13 joules
Momentum =   5.63918E-22 kgm/s pc= 1.6906E-13 joule
1.05664076 MeV
or expressing the electron mass in energy equivalent unit 0.511 MeV/c^2
p= gamma*m*.9c=  1.05508296 MeV/c

non rel ke=  3.31617E-14 joules
207260.5635 eV 0.20726056 MeV
error percent=  68.6592998 %
non rel momentum=  2.45806E-22 error 129.415734 %
in energy equivalent units=  0.4599 MeV/c

5. Sodium photoelectric effect
a. Maximum wavelength = hc/2eV=   6.20586E-07 meters
b. KE of electron = hc/wavelength - 2 eV=   1.10293 eV

6. A. -13.6*(1/25-1/4)  2.856 eV
Wavelength = hc/E =  4.34584E-07 m

b. energy = (42-1)^2*13.6*(1 - 1/4)=   17146.2 eV
Wavelength = hc/E =  7.23876E-11 m

7. Lava blob
a. Radiated power from Stefan-Boltzmann law
area= 4 pi r^2=  3.141592654 m^2
temperature=  3000 K

b. cooling rate from dE = N3k/2 * dT
and dE/dt = sigma*area*T^4
This gives an integral for time   dt = dE/(sigma*A*T^4) =(3Nk/(2*sigma*A))*dT/T^4
The cooling time is then t=   Nk/(2*sigma*A)*(1/Tf^3 - 1/Ti^3)

c. Assuming a trajectory time back to the height of the launch
t = 2*100m/s/9.8 m/s^2=  20.40816327 seconds
Rearranging the cooling equations gives
1/Tf^3 = 2*sigma*A*t/Nk + 1/Ti^3
We need a value for N, so some assumptions must be made
Assume density of 3 gm/cm^3, mass 40 amu
Mass of lava =   1570.796327 kg
Number of atoms/molecules =  2.36566E+28  =N
The first term in the solution above is then =   2.22602E-11 1/K^3
The second term in the expression = 1/Ti^3 =   3.7037E-11 1/K^3
The resulting final temperature from inverting and taking cube root=     2564.41677 K
If the peak were 1000 m high and we neglect air friction the final temperature becomes
t= 27.75984748 sec
first term above gives   3.0279E-11
temperature upon impact =  2458.256312 K so it doesn't make that much difference
So with only radiative cooling, it might still be molten when it hits the poor suckers on the slope below.

8. Cloud chamber track is 5 mm long
a. The gamma factor is =5, so the speed is given by v/c= sqrt(1 - 1/25)=     0.9797959
b. The lifetime is distance in its frame divided by the velocity, upon which both observers agree.
Particle frame sees length contracted to 1 mm
Time is 1mm/.98c =  3.33556E-12 seconds

9. Muon experiment
a. Time of transit = 1600/c  5.33689E-06 seconds
b. 422/525=ratio  0.803809524
-.693T/halflife = ln(ratio) gives halflife    1.69349E-05 seconds
this gives a gamma factor of  10.85571364
this gives a speed =v/c=sqrt(1-1/gamma^2)=   0.995748154