#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.