Homework 3, Physics 3401, Chapter 3,#10,19,22,37  Rohlf plus handout.

#10 Maximum visible distance for sun-type star.
Assume visible if 250 visible photons per sec enter the eye
Radius of pupil 2.00E-03 m
Solar luminosity from pg 74 of Rohlf   3.83E+26 watts
For an assumed solar temperature of 5700K, the peak is at a quantum energy
given by Wien's displacement law as    2.44 eV
Really need to know what fraction of energy is in the visible, but I don't know
that right now, so approximate by assuming all in the visible.
Convert photon energy to joules:  3.90E-19 J/photon
e = 1.60E-19 j/eV
Number of photons:  9.81E+44 photons/s
Area of pupil=pi r^2 =   1.26E-05 m^2
Photon flux required = 250/area=  1.99E+07 photons/m^2 second
Maximum distance is radius of sphere where this flux would sum to the number of
photons from the sun calculated above.
R =sqrt(total # of photons/(required flux*4*pi))   1.98E+18 meters

#19 Energy density of cosmic background radiation.
The average radiated energy given by the Stefan-Boltzmann relationship is c/4 times the
energy density.
Radiated energy per unit area in watts/m^2 at 2.7K =    3.01E-06 W/m^2
The energy density at temperature = 2.7K is then 4/c times that figure:
Energy density = 4.01E-14 joules/m^3

#22 Energy of photons from different sources
Photon energy = h * frequency
h =  4.14E-15 eV s
a. FM radio at 100 MHz, photon energy =   4.14E-07 eV
b. Microwave overn o.o1 meter wavelength =   1.24E-04 eV
c. The Sun, assume 5700K, use Wien   2.44E+00 eV
d. Ceramic at 1000K from Wien displacement peak   0.428 eV
e. Cosmic background at 3K from Wien   1.28E-03 eV
Note that the cosmic background is most comparable to the microwave oven

#37 Maximum wavelength aborbed by electron in its first excited state
The minimum energy transition from n=2 is to n=3, so the photon energy
is 13.6eV(1/4 -1/9) =   1.889
The corresponding wavelength is   656.8 nm when 13.6eV is used.
This is the familiar red line of hydrogen.

Handout assignment
#1  Number of photons from your index finger
Finger area and temperature needed
Take temperature T= 34C =   307 K
Assume finger to be cylinder 8.5 cm, 1 cm radius
Surface area, neglecting ends, is pi*1^2*8.5cm*1e-4 =    0.002670354 m^2
Assuming e=1 (perfect emitter) and external temperature 0K
to get just the radiated power.
Power radiated = (5.67e-8*2.67e-3*307^4) =    1.344947584 watts
Wien peak = 2.898e-3mK/307.15K =    9435 nm
Energy associated with this characteristic photon = 1240 eV nm/9435nm=     0.1314
k 1.38E-23 J/K
Average photon energy kT =   4.2386E-21 joule
Number of photons = 1.34J/s/(1.6e-19J/eV)(0.131eV)=    6.39596E+19 photons/second

#2 60 watt light bulb
Filament temperature 2780K
Area = 60W/((5.67e-8 W/m^2K^4)*2780K^4) =   1.7717E-05 m^2
Wien peak is = 1042.4 nm
Energy of photon = 1240 eV  nm/1042.4 nm =   1.19 eV

The above figures were obtained by setting the ambient temperature equal to 0 K to get the output.
Taking into account some re-radiation from the environment and calculating the net output
for a 300K environment made no difference to this level of precision because the temperatures
are raised to the fourth power, making 300K insignificant compared to 2780K.

1.60E-19 J/eV
so for 60J/s, the required number of photons is    3.15E+20 photons/s

From Javascript calculation, in the range 400-700 nm P=3.42 watts or 5.7% of the total radiation.

#3 Hot potato
Take hot potato temperature= 100 C = 373K
Cool potato temperature = 50C = 323K
Taking the potato to be sphere of radius 4cm and density 1.2 g/cm^3 gives mass 322 gm
Guess the mean molecular mass to be 0.8*18 (water) + 0.2*200(something organic) = 54

Tcooling = (Nk/2*sigma*A)*(1/Tc^3-1/Th^3) =   226 s
Using N=3.59e24 and A=0.02 m^2

#4 Hot rock For basalt at 3 g/cm^3, mass 804 gm, guess same molecular weight
Tcooling from 1673K to 1473K = 5.4 s.
For 2.7 sec to peak, the launch velocity is approximately 27 m/s.

#5 Body cooling
For 2 m^2 body at 34C radiating into room at 22C, the radiative loss rate is 149 watts.

Revising insulator thickness to 1cm with thermal conductivity k=0.04 W/mK gives a
conduction loss rate of
Q/t=kAdeltaT/thickness = (0.04W/mK)*(2m^2)*12C/0.01m =    96 watts

Forcing the heat loss to be conduction and then radiation gives a series loss equation

kA(Tbody -Tsurface)/thickness = sigma*A(Tsurface^4-295K^4)
.04*2*(307-T)/.01=sigma*2*(T^4-295^4)
Iterating these two expressions below gives a surface temperature of 299.82 K
and a heat loss rate of 57.5 watts.
300            56          59.722
299.98       56.16    59.477
299.96       56.32    59.232
299.94       56.48    58.987
299.92       56.64    58.742
299.9         56.8      58.497
299.88       56.96    58.253
299.86       57.12    58.008
299.84       57.28    57.764
299.82       57.44    57.519
299.8         57.6      57.275
299.78       57.76    57.030

#6 Energy content of body.
Using N3/2(kT) at 37C with density of body =1, using 18 for molecular weight (water)
200 lb = 90.78kg, 90.78*6.02e23/.018 =   3.03609E+27 molecules

E=3e27molecules*3/2*1.38e-23*(273+37)=   19482568.14 joules
E= 4654. Dietary Calories (kcal)

#7 Photoelectric effect for sodium with work function 2.3 eV
Wavelength Photon energy hf-W Velocity
640 1.94    ***    ***
530 2.34   0.04   118701.
440 2.82   0.52   427983.

#8 K-alpha xray energy for molybdenum, Z=42

E =13.6*(42-1)^2*(1-1/4)=  17146.2 eV

#9 For cesium, the 6s electron ionization energy is 3.89 eV.
The target photon energy for f=9GHz is 0.372e-4 eV
Taking hydrogenic energy expression, 13.6 Zeff^2/6^2 = 3.89 eV
This gives an effective atomic number Zeff = 3.209
The target energy level pair meets the condition:
13.6*(3.209^2)*(1/n^2-1/(n+1)^2)=0.372e-4
Approaching the energy level by iteration:

n transition energy  3.72E-05
195     3.74864E-05
196     3.6917E-05
197     3.63591E-05
198     3.58123E-05
199     3.52765E-05

So it would take n=195 for straight hydrogenic levels, which is physically
unrealistic. This suggests that the actual level separation used in the
cesium clock is some other kind of level difference - some kind of
fine structure.