The Proton-Proton Chain is the principal set of reactions for
solar-type stars to transform hydrogen to helium:
^{1}H + ^{1}H --> ^{2}H + e^{+} + neutrino
Two protons react to form deuterium plus a positron and a neutrino
^{2}H + ^{1}H --> ^{3}He + gamma-ray
The deuterium reacts with another proton to form ^{3}He plus another
gamma-ray. ^{
3}He + ^{3}He --> ^{4}He + 2 ^{1}H
Two ^{3}He nuclei react to form ^{4}He and two protons.
Each proton-proton cycle generates 26.7 MeV of energy. The Sun emits
approximately 4*10^{26} Watts of energy. Calculate the rate at which
the sun generates neutrinos and estimate the number of solar neutrinos aping
through Earth each second. Assume the sun's energy production is entirely by the
proton-proton cycle.
Earth-Sun distance: 1.5*10^{-11} m
Earth radius: 6.37*10^{6} m
Solution:
Most of the neutrinos from the Sun are produced in the chain of processes
called the “pp-cycle”:
(1)^{ } p + p --> ^{2}H + β^{+} + ν
(2)^{ } p + ^{2}H --> ^{3}He + γ
(3) ^{3}He + ^{3}He --> ^{4}He + p + p
Estimate the order of magnitude of the neutrino flux (neutrinos/(cm^{2}s)
from these reactions on Earth using the following data:
Earth-Sun distance = 1.5*10^{11} m
proton mass = 938.272 MeV/c^{2
4}He mass = 3727.379 MeV/c^{2}
Assume the Earth can be modeled as a black body with temperature T = 300 K. On
average, it emits as much radiation as it receives from the Sun. Use this to
estimate the energy flux from the sun on Earth.
Solution:
What is the solar energy flux at the sun-earth distance?
We assume that it is equal to the power per unit area emitted by Earth.
I = σT^{4} = (5.67 * 10^{-8} W m^{-2} K^{-4})*(300
K)^{4} = 459 W/m^{2} = 2.87*10^{15} MeV/(m^{2}s).
We therefore expect a neutrino flux of approximately 2* 2.87*10^{15}
/25.7 neutrinos/(m^{2}s),
or 2.23*10^{14} neutrinos/(m^{2}s) = 2.23*10^{10}
neutrinos/(cm^{2}s).
(The energy flux is underestimated, the real neutrino flux is ~6*10^{10}
neutrinos/(cm^{2}s).)
The Wien law and the Stefan-Boltzmann law
The light from the sun is found to have a maximum intensity near 470 nm. Assuming the surface of the sun behaves as a black body, calculate the temperature of the sun.
Solution:
An ideal radiator radiates
with a total intensity of I = 5.68 kW/m^{2}. At what wavelength does
the spectral emittance I(λ) peak?
(Give a numerical answer!)
Solution:
Two concentric long tubes have radii R_{1} = 5 cm and R_{2} = 6 cm. The outer tube is maintained at temperature T_{2} = 300 K and the inner tube at temperature T_{1} = 4 K. Find the net thermal power absorbed by a 10 cm length of the inner tube, assuming that the emissivity is unity for both surfaces.
Solution:
Estimate the temperature of the surface of Earth if the flux of solar energy at the Sun-Earth distance is ~1360 W/m^{2 }and ~30% of solar energy is reflected back by the atmosphere. (Make reasonable assumptions and justify them.)
Solution:
A black plane surface at a constant high temperature T_{h} is
parallel to another black plane surface at a constant lower temperature T_{l}.
Between the plates is vacuum.
In order to reduce the heat flow due to radiation, a heat shield consisting
of two thin black plates, thermally isolated from each other, is placed between
the warm and the cold surfaces and parallel to these. After some time
stationary conditions are obtained. By what factor X is the stationary
heat flow reduced due to the presence of the heat shield? Neglect end
effects due to the finite size of the surfaces.
Solution:
A sphere of radius 3 cm acts like a blackbody. It is equilibrium with its surroundings and absorbs 30 kW of power radiated to it from the surroundings. What is the temperature of the sphere?
Solution:
The rate at which the radiant energy reaches the surface of earth from the
sun is about 1.4 kW/m^{2}.
The distance from earth to the sun is
about 1.5 * 10^{11} m, and the radius of sun is about 0.7 * 10^{9}
m.
(a) What is the rate of radiation of energy, per unit area, from the
sun's surface?
(b) If the sun radiates as an ideal black body, what is the
temperature of its surface?
Solution:
In very massive stars the pressure from electromagnetic radiation can greatly
exceed that from the gas.
(a) Show that in the hot interior of a star
the radiation pressure is P_{rad} = aT^{4}/3, where
a = 4σ/c
and σ is the Stefan Boltzmann.
(b) Assuming a spherical star described
by classical Newtonian gravity and a classical electromagnetic field, find an
expression for the temperature gradient at the surface of a star when the force
per unit volume associated with the radiation pressure is exactly balanced by
the force per unit volume associated with gravity, assuming that the gas
pressure can be neglected.
(c) For stars dominated by radiation the
temperature gradient is known to take the form
dT/dr = -3ρkL/ (16π a c T^{3}r^{2}),
where T is temperature, L = L(r) is the luminosity (energy per unit time
crossing the radius r), and k is a measure of how strongly the photons interact
with matter (opacity). Use this and the first result to show that the star
is unstable against radiation blowing surface layers off the star if the
luminosity exceeds L = 4πcGM/k, where k is evaluated at the surface.
Hint:
You can complete parts (b) and (c) without completing part (a) by using the
result given in part (a).
Solution:
The probability that this energy will reach
the area dA is dA cosθ/(4πr^{2}).
Denote the photon energy
density in the ring by u.
The amount of energy E reaching dA from the
ring is
u2πr^{2}sinθ dr dθ dA cosθ/(4πr^{2}) = u sinθ dr
dθ dA cosθ/2.
The total energy reaching dA in a time interval dt comes
from rings with radii between r = 0 and r = cdt. It is therefore given
by
I dA dt = ∫_{0}^{π/2} dθ u dA cdt sinθ cosθ/2 = ¼
u dA cdt.
I = ¼ u c.
Setting (u c)/4 = σ*T^{4}, we find the energy density in the
volume u = (4σ/c)*T^{4}.
The pressure exerted by a photon
gas is P = u/3.
So P_{rad} in the interior of the star is
(4σ/(3c))*T^{4} = (a/3) T^{4}.
(For any gas: PV = N<p∙v>/3.
For a photon p = E/c and v = c, so p∙v = E.
N<E> = U = internal
energy of the photon gas, P = U/(3V) = u/3.)
(b) Consider a thin
spherical shell of radius r and mass dm = ρ4πr^{2}dr inside the
star. The gravitational force per unit area on this layer is is Gm_{inside}
ρdr/r^{2}, pointing towards the center of the star. In
equilibrium, P_{top} + Gm_{inside} ρdr/r^{2} = P_{bottom}.
dP/dr = -Gm ρ/r^{2} near the surface of the star. If the
star is supported by radiation pressure P_{rad} = aT^{4}/3,
then we need dP_{rad}/dr = (4/3)aT^{3}(dT/dr) = -Gm ρ/r^{2}.
We need dT/dr = -3Gm ρ/(4aT^{3}r^{2}).
(c) Let dT/dr
= -3ρkL/ (16π ac T^{3}r^{2}) = -3Gm ρ/(4aT^{3}r^{2}),
i. e. let the star be in equilibrium.
Then L = 4πcGm/k.
If L
exceeds this value the surface layer of the star experiences a net outward
force.
The Sun's spectrum, I(hν), peaks at 1.4 eV, the spectrum of Sirius A peaks at 2.4 eV, and the luminosity (total amount of energy radiated per unit time) of Sirius A is 24 times larger than that of the Sun. Compute the diameter of Sirius A in units of the Sun's diameter.
Solution:
Sirius A is the brightest star in the
night sky, with the peak of its spectral emittance at a wavelength of 291
nanometers.
(a) If one makes the reasonable assumption that the star radiates as a
blackbody, what is its effective temperature?
(b) If the flux measured on Earth from Sirius A is 1.17 * 10^{-7}
W/m^{2}, and the distance is known to be 8.6 light-year, what is the
radius of Sirius A?
Solution:
Planck's law
Inside a blackbody cavity, the energy density per unit frequency interval, ρ(ν), is given by Planck's formula
ρ(ν) = (8πν^{2}/c^{3})hν/(exp(hν/(kT)) - 1).
(a) Derive an expression for the intensity per unit frequency interval, I(ν),
of the radiation emitted by the blackbody.
(b) Derive the Stefan-Boltzmann law.
(c) Derive Wien's displacement law.
∫_{0}^{∞}x^{3}dx/(e^{x} - 1) = π^{4}/15.
Solution:
The probability that this energy will
reach the area dA is dA cosθ/(4πr^{2}).
The amount of energy per unit frequency interval reaching dA from the ring
is
ρ(ν)2πr^{2}sinθ dr dθ dA cosθ/(4πr^{2}) = ρ(ν)sinθ dr dθ dA
cosθ/2.
The total energy per unit frequency interval reaching dA in a time interval dt comes from rings with radii between r = 0 and r = cdt. It is
therefore given by
I(ν) dA dt = ∫_{0}^{π/2} dθ ρ(ν) dA cdt sinθ cosθ/2 = ¼
ρ(ν) dA cdt.
I(ν) = ¼ ρ(ν) c.
(b) To find the total intensity, integrate over all frequencies.
∫_{0}^{∞}(2π/c^{2})hν^{3} dν/(exp(hν/(kT)) -
1) = (2πk^{4}T^{4}/(h^{3}c^{2}))∫_{0}^{∞}
x^{3} dx/(exp(x) - 1)
= (2πk^{4}T^{4}/(h^{3}c^{2}) π^{4}/15
= (2π^{5}k^{4}T^{4}/(15h^{3}c^{2}) =
σT^{4}.
Here σ is the Stefan-Boltzmann constant.
(c) The Wien Law gives the wavelength of the peak of the radiation
distribution.
|I(ν)dν| = |I(λ)dλ|, I(λ) = I(ν)|dν/dλ| = (c/λ^{2})I(ν).
I(λ,T) = [2hc^{2}/λ^{5}][1/(exp(hc/(λkT)) - 1)].
dI(λ)/dλ = 0 --> (d/dλ)(λ^{5}(exp(hc/λkT – 1))^{-1} = 0.
[hc/(λkT)][[exp(hc/(λkT))/(exp(hc/(λkT)) - 1)^{2}] - 5 = 0,
xe^{x}/(e^{x} - 1) - 5 = 0.
Use a calculator to find x ~ 4.96.
λ_{max} = hc/(xkT) ~ (2.9*10^{-3} mK/T).
Inside a blackbody cavity, the energy density per unit frequency interval,
ρ(ν), is given by Planck's formula
ρ(ν) = (8πν^{2}/c^{3})
hν/(exp(hν/kT) - 1).
The intensity per unit frequency interval, I(ν), of the
radiation emitted by the blackbody is given by I(ν) = ¼ ρ(ν) c.
Commonly,
Wien's displacement law is written as λ_{max} (m) = (2.9*10^{-3}
m K)/T.
Derive the Wien's displacement law for the frequency ν_{max}
(s^{-1}), and show that ν_{max} is NOT equal to c/ λ_{max}.
Why?
Solution: