Homework 3, Physics 3401, Chapter 3,#6,7,8,10,14,17,19,22,25,36,37,40,44  Rohlf

#6 Number of photons from your pencil
 Pencil area and temperature are needed, and avg energy per photon
 Take temperature T=  300 K
 Measured pencil to have length 19 cm and diameter 0.8 cm
 Surface area of pencil, neglecting ends is    47.75220833 cm^2
 Assuming e=1 (perfect emitter) and external temperature 0K
 to get just the radiated power.
 Power radiated =   2.19 watts
 Average photon energy of kT assumed.
 k 1.38E-23 J/K
 Average photon energy kT =   4.14198E-21 joule
 Number of photons =   5.28733E+20 photons/second
#7 Photon energy at the peak of the solar spectrum.
 Take the temperature of the sun to be 5700K and use Wien displacement law
 This gives a peak at 508 nm and at photon energy 2.44 eV
 But this is the peak with respect to wavelength.
 From average thermal energy 3kT/2 = E we get an average photon energy of 0.74 eV
 Peak with respect to frequency approx 3kT =    1.473507 eV
 k 8.62E-05 eV/K
 By interation, real max is at 2.82KT =   1.38509658 eV
 So Rohlf's 3kT approximation is really not all that useful
#8 100 watt light bulb
 Assume blackbody spectrum with white output
 This puts peak wavelength at about 500 nm (Sun is a little yellow with peak 550)
 Wien displacement law gives temperature of
  Constant = 0.002898
  T=Constant/wavelength =  5796 K
 Stefan-Boltzmann law gives area:  1.56E-06 m^2 = 0.0156 cm^2
 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 5796K.
 To get an estimate of the number of photons, the Wien displacement law gives a photon energy
 of  2.48 eV
  1.60E-19 J/eV
 so for 100J/s, the required number of photons is    2.52E+20 photons/s

#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))   1.98095E+18 meters

#14 The Kelvin cooling time for the earth.
 The earth is now at 300K but was once in a very hot molten state.
 The cooling rate is given by the Stefan-Boltzmann law
 The thermal energy of the earth is the number of particles times 3kT/2, and the
 rate of cooling could be expressed as a derivative of that energy:
 Change of energy = dE = N*3k/2*dT
 Since the Stefan-Boltzmann equation gives dE/dt, we can set up and expression for
 the differential of time in terms of temperature and then integrate it.
 The Stefan-Boltzmann equation is of the form C/T^4, so the integrrand  for dt is of the form
 C dT/T^4, giving an expression in terms of T^3 when integrated.
 The term involving the hot initial temperature can be neglected since it is in an inverse
 cube relationship. The physical rationale for that is that the cooling rate is very fast then,
 and that part of the cooling time is very small compared to the slower cooling as it approached
 the current earth temperature.
 Redefine C to be the constant after integration, which has a factor of 3 in denominator from integration
 The constant C above has N*3k/2 in the numerator and 3*Earth area * Stefan Boltzmann constant in
 the denominator.
 To estimate the number of atoms in the earth, we can use the nominal atomic volume from Rohlf:
 Volume of atom =  1.00E-29 m^3
 Radius of earth = 6.40E+06 m
 Volume of earth= 1.10E+21 m^3
 Number of atoms in earth = N =  1.10E+50 atoms
 k 1.38066E-23 J/K
 Numerator of fraction = N3k/2 =  2.27E+27 J/K
 Area of earth 4 pi r^2 =  5.15E+14 m^2
 Stefans constant=  5.67E-08 W/(m^2 K^4)
 The constant C is then =  2.60E+19 seconds/K^3
 Dividing by the cube of 300K gives time =   9.62E+11 seconds
   Time in years 30483.48924 years

#17 Naked person at South Pole
 The heat loss from a person by radiation is governed by the Stefan-Boltzmann law, and
 the standard statement is that about 2/3 of the energy loss is by radiation. Since we are
 estimating, assume all loss is by radiation.
 Another nominal figure is that the room temperature loss from a person at room temperature
 is about 90 watts on a 24 hour basis. If assume 34 C as skin temp and 22 C as room temperature,
 this gives an effective radiating area from Stefan-Boltzmann as Area =    1.2 m^2
 Assume South Pole temperature =   210 K
 Body temperature given =  311 K
 Assuming ideal radiator (actually about 95%) then we get a loss rate of:
 Power radiated = 509 watts  compared to nominal 90 watt loss rate.
 If the external temperature is at   290 K
 The loss rate is only   156 watts
 Assuming a body mass of 80 kg and an average mass of atom of 18*1/.8 since we are 80% water,
 u  1.66E-27 kg
 Mass of atom in body =  3.735E-26 kg
 Number of atoms in body = N =   2.1419E+27 atoms
 Average energy of those atoms = N3kT/2
 k 1.38066E-23 J/K
 At 311K, N3kT/2 =  13795510.36 joules
   3295.63076 dietary Calories

#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

#25 Wavelength of incident radiation, photoelectric effect
 Work function = 2.3 eV, photoelectron energy 0.7 ev, incident photon energy 3 eV
 Corresponding wavelength =  414 nm, in the blue visible

#36 Minimum wavelength emission from hydrogen atom
 The minimum wavelength photon, max energy would occur for a free electron
 dropping straight in to the ground state at -13.6 eV
 That corresponds to a frequency of 3.29 E15 and a wavelength of 91.2 nm in the uv.

#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) =   1.888888889
 The corresponding wavelength is   656.8 nm when 13.6eV is used.
 This is the familiar red line of hydrogen.

#40  Hydrogen and helium emissions
 The 3->2 transition is computed to be 656.8 nm if 13.6 eV ionization energy is used
 Tne nuclear charge goes into the Bohr orbit energies as a square, so the singly ionized
 helium ground state is at -4*13.6 = -54.4 eV. The levels are -54.4, -13.6,-6.04, -3.4, -2.18, -1.51
 Examination shows that the 6->4 transition of helium is the same as the 3-2 of hydrogen to
 this level of approximation.

#44  Lead K-alpha x-ray
 For lead, Z=82, but the effective Z for the K-alpha x-ray is 81 since we presume that there
 is one electron in the n=1 shell so that the nucleus is shielded by that electron.
 Other than the shielding, the energy levels are presumed to be like those of hydrogen
 and the n=2 to n=1 transition is then = (13.6*81^2*(1-1/4) =     66922.2 eV
 So we expect about a 67 keV xray.