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Flux and Luminosity
The luminosity L of a star etc is simply its total power output in watts (W). For
example, the luminosity of the Sun is 3.84 x 1026 W.
Stars are approximately black bodies and therefore emit black body radiation.
That is to say, they emit electromagnetic radiation of all wavelengths (radio gamma), but their output peaks at some particular wavelength. The Sun's output
peaks in the visible region. (The temperature of the surface of the Sun is
approximately 6000 K.) The output of a much hotter star would probably peak in the
UV region, and that of a much cooler star in the IR region. Thus, the Sun appears
yellowish, a hotter star would appear bluish, and a cooler star would appear reddish.
The flux F from a star etc is the power per unit area received at some point a
distance d from the star. The units of flux are therefore Wm-2. This is illustrated by
the diagram below .....
At a distance d from the star .....
flux F = power per unit area = L/(surface area of sphere of radius d)
Thus .....
F = L/4d2
If the luminosity of a star is known and its flux can be measured, the above formula
can be used to calculate its distance d.
Nuclei and Isotopes
Protons and neutrons are collectively known as nucleons.
A nuclide is a nucleus of a particular type - with a particular number of protons and
neutrons.
The mass number or nucleon number A of a nuclide is the total number of
nucleons in it - i.e. the total number of protons and neutrons.
The proton number or atomic number Z of a nuclide is the number of protons in it.
For a neutral atom, this is the same as the number of electrons surrounding the
nucleus. Thus Z of the nucleus determines the chemical properties of the atom.
Nuclides with the same atomic number Z but different numbers of neutrons (and
therefore different mass numbers A) are known as isotopes of that element. (Note
that they all have the same chemical properties because they have the same
number and arrangement of electrons.) This is illustrated below .....
Some isotopes (e.g. carbon 12) are stable. Others (e.g. carbon 14) are unstable
and radioactive. i.e. They emit alpha, beta or gamma radiation.
Radioactive Decay
An unstable isotope (the parent nuclide) may emit an alpha or a beta particle and
decay to form an isotope of another element (the daughter nuclide), as shown
below .....
Note that both the mass numbers (top line) and the charge (bottom line) balance.
Gamma radiation may often accompany the above. No change of A or Z is caused
by gamma emission, since gamma particles (photons) have neither mass nor
charge.
Radioactive decay is a random process. This means that it is impossible to tell
when exactly a given parent nucleus will decay into a daughter nucleus. All we can
say is that there is a certain probablility that the nucleus will decay in unit time
(usually 1 second).
This probability (per unit time) is called the decay constant . It is a constant for a
given unstable isotope, but varies greatly from isotope to isotope depending on the
half life (see below). Its units are s-1, min-1, year-1 etc.
We can also say that the decay constant  is the fraction of the parent nuclei
decaying in unit time.
Thus .....  = - dN/N / dt
..... where dN/N is the fraction of the parent nuclei decaying in time dt. ("N" stands for
the number of parent nuclei left at any given time. "dN" is the change in that number.
The minus sign is there because dN is negative (N is decreasing), and we want  to
be positive.)
Thus .....
dN/dt = - N
dN/dt is the rate of change of N. i.e. It is the number of parent nuclei decaying per
second. This is also equal to the number of alpha (or beta) particles emitted per
second. Another word for this is the activity A of the radioactive source, measured
in becquerel (Bq). (1 Bq = 1 particle emitted per second.)
If we integrate the above equation, we obtain a relationship between the number of
parent nuclei remaining (N) and time (t) .....
N = N0e-t
(N0 is the number of parent nuclei present at t = 0.)
Graphs of N against t and logeN against t are shown below .....
The above is an example of exponential decay.
Since N = N0/2 at t = t1/2 (the half life), it follows that .....
N0/2 = N0e-t1/2
Therefore ..... e-t1/2 = 1/2
Therefore ..... e+t1/2 = 2
Therefore ..... t1/2 = loge2
Therefore .....
 = loge2 / t1/2
i.e. The longer the half life (t1/2), the lower the probability of decay in unit time () and vice versa.
Nuclear Fission and Fusion
Fission involves splitting a heavy nucleus (typically 235U), by bombarding it with a
neutron. 2 lighter nuclei are formed, plus some more neutrons which go on to cause
further fission - and hence a chain reaction. Energy is released in the form of kinetic
energy (heat), which can be used to raise steam which drives the turbines and
generates electricity in a nuclear power station. A typical fission reaction is shown
below .....
Fusion is the source of the energy that stars (such as the Sun) emit. In fusion,
lighter nuclei join together to form a heavier nucleus. (The most basic fusion reaction
involves hydrogen nuclei fusing to form helium.) As with fission, kinetic energy (heat)
is released as well as photons. A typical fusion reaction is shown below .....
In order to make fusion happen you need very high temperature (typically about
107 K) and pressure. High temperature is needed in order to give the hydrogen
nuclei enough velocity (and hence k.e.) to overcome their electrostatic repulsion.
(They are positively charged.) High pressure is needed to produce a high density of
particles, and hence a great enough probability of collision and reaction.
The graph below can be used to explain why energy is released in both fission and
fusion .....
The binding energy is the energy that holds the nucleons (protons and neutrons)
together in a nucleus. It is negative because nucleons are thought of as having zero
potential energy when they are an infinite distance apart, and they lose potential
energy as they "fall" closer togther. i.e. Their potential or binding energy becomes
negative. The binding energy per nucleon is the total binding energy divided by the
number of nucleons. The more negative the binding energy per nucleon the more
tightly the nucleus is bound together. (Iron has the most negative BE per nucleon
and is therefore the most stable nucleus.)
(N.B. We sometimes refer to binding energy as though it were positive. In that case,
we are really talking about the positive amount of energy that it would take to pull the
nucleons apart - the "unbinding" energy. Thus we might say that iron has the
greatest binding energy per nucleon, even though it is actually the most negative.)
Note that in both fission and fusion the binding energy per nucleon after the reaction
is more negative than it was before the reaction. This fall in one kind of energy
results in an increase in other kinds of energy. Thus the reaction products have
greater kinetic energy than the nuclei etc that were present before the reaction. Heat
is produced as well as photons.
In any reaction where the binding energy (E) becomes more negative there is also
a loss of mass (m). i.e. The reaction products have less mass than the particles
that go into the reaction. (This applies also when nucleons are brought together to
form a nucleus.) This loss of mass is known as the mass defect. The relationship
between E and m is .....
E = c2m
Thus the mass defect can be used to calculate the energy released in fusion or
fission - or the total binding energy of a nucleus.
For example, the total mass of the barium and krypton atoms plus the 3 neutrons in
the above fission reaction is 3.090 x 10-28 kg less than the mass of the uranium
nucleus plus 1 neutron. Using E = c2m, this means that the energy released in the
reaction is 2.78 x 10-11 J. The energy released per kg of uranium could then be
calculated if we knew the number of atoms of uranium (and hence the number of
reactions) per kg.
The Hertzsprung-Russell Diagram
A Hertzsprung-Russell diagram (or HR diagram for short) is essentially a graph of
luminosity against temperature for a sample of stars. Each star is a point on the
graph. It is a way of displaying information about stars and showing trends and
patterns. The scales are usually log-log to cope with the wide range of values
(particularly for luminosity) and, by convention, the temperature scale goes
backwards - i.e. high temperatures are on the left of the axis.
A typical HR diagram for a large population of stars is shown below .....
The majority of stars (including the Sun) are in the main sequence - a line which
runs from massive, luminous, hot stars at one end to low mass, dim, cool stars at the
other end. Another group of stars, the red giants, are relatively cool - but they are
very luminous, because their diameters and surface areas are very large compared
with main sequence stars. The white dwarfs are hot but not very luminous because their diameters are very small. Supergiants are very large and luminous,
and their temperatures cover the full range from very hot to relatively cool.
An HR diagram is really a snapshot of many stars at different stages in their lives.
(The process of change that occurs throughout a star's life is known as stellar
evolution.) The reason there are so many stars on the main sequence is that stars
spend most of their lives there. They start their lives away from the main sequence,
but it only takes a relatively "brief" (in astronomical terms) time for them to evolve
onto the main sequence. Only towards the end of their lives do they "briefly" evolve
through other areas of the diagram such as the red giant and white dwarf regions.
More massive stars evolve more rapidly and have briefer lives than less massive
stars, because they are are hotter and fusion proceeds more rapidly and violently.
The most massive stars will end their lives in supernova explosions.
Gravitation
It was Newton who first realised that there is a force (F) of attraction between all
pairs of masses in the universe, which is proportional to the product of their masses
(m1 and m2)and inversely proportional to the square of the distance r between them,
as shown below .....
This gravitational force F is given by the equation .....
F = Gm1m2 / r2 REMEMBER!
..... where G is a universal constant called the gravitational constant. G = 6.67 x
10-11 Nm2kg-2.
Gravitational field strength g is defined by the equation .....
g = F/m
..... where F is the force on a mass m placed in the field. The units of g are Nkg-1.
Suppose the field at the mass m is due to another mass M which is a distance r
away. The force F on m due to M's field will be given by .....
F = GMm / r2
Therefore the gravitational field strength g a distance r from M is given by .....
g = F/m = (GMm / r2)/m = GM / r2
g = GM / r2
Many of the equations for gravitational and electric fields are similar. The table
below summarises some of the similarities and differences between the 2 types of
field .....
Gravitational fields
Electric fields (see
PRO)
act on matter
act on charges
forces are always
attractive
forces can be
attractive or repulsive
force law formulated
by Isaac Newton
force law formulated
by Charles Coulomb
F = Gm1m2 / r2
F = kQ1Q2 / r2
field strength g =
F/m Nkg-1
field strength E =
F/q NC-1 or Vm-1
The direction of the
field g is the same as
the direction of the
force - i.e. towards a
mass.
The direction of the
field E is the same as
the direction of the
force on a positive
charge - i.e. away from
a positive charge and
towards a negative
charge.
Ideal Gases
An ideal gas is one which obeys the ideal gas equation (or the equation of state
for an ideal gas) .....
pV = nRT REMEMBER!
..... where p is the pressure of the gas in Nm-2, V is its volume in m3, n is the
number of moles of the gas, R is the universal gas constant (8.31 Jmol-1K-1) and T
is the temperature of the gas in K.
Real gases do not obey the above equation exactly - particularly at very low
temperatures and high pressures. Under these conditions they may even liquefy - in
which case ideal gas behaviour clearly breaks down. (The particles of an ideal gas
would have zero volume, for example. Although the atoms/particles of a real gas are
small compared with their distances apart, their volumes are not totally negligible.)
Re-arranging the ideal gas equation, we obtain .....
pV / T = nR = a constant
or ..... p1V1 / T1 = p2V2 / T2
If T is also a constant we have .....
pV = a constant
or ..... p1V1 = p2V2 ..... or ..... p  1/V
which is Boyle's Law
If p is a constant we have .....
V / T = a constant
or ..... V1 / T1 = V2 / T2 ..... or ..... V  T
which is Charles's Law
If V is a constant we have .....
p / T = a constant
or ..... p1 / T1 = p2 / T2 ..... or ..... p  T
which is the Pressure Law
Graphs for the above 3 gas laws are shown below .....
Absolute zero (0 K) is defined as the temperature at the which the pressure (or
volume) of an ideal gas becomes zero. (See the Charles's Law and Pressure Law
graphs above.)
absolute zero = - 273oC
temperature in K (absolute temperature) = temperature in oC + 273
A real gas would, of course, liquefy and solidify before its temperature reached
absolute zero - so its volume could never become zero.
The greater the absolute temperature T, the greater the average kinetic energy of
the particles of a gas.
In fact ..... average ke of particles  T ..... [1]
The symbol "c" rather than "v" is used for the velocity of a single particle in a gas.
Therefore the kinetic energy of a single particle of a gas is given by .....
ke of particle = 1/2mc2 ..... where m is the mass of the particle/atom/molecule
But c varies from particle to particle, as shown below .....
Therefore the average ke of the particles in a gas is given by .....
average ke of particles = 1/2m x (average value of c2)
This is usually written .....
average ke of particles = 1/2m<c2> ..... where "< >" means "average value of"
Substituting into equation [1] above .....
/2m<c2>  T
1
The constant of proportionality in the above equation is 3/2k, where k is Boltzmann's
Constant (k = 1.38 x 10<-23 JK-1).
Therefore .....
1
/2m<c2> = 3/2kT
Note that ..... "<c2>" is not the same as "<c>2" ..... ! <c2> is also called the mean
square speed of the particles. The root mean square speed crms is the square root
of the mean square speed. Thus we could also write .....
/2mcrms2 = 3/2kT
1
Note also that ..... the average ke of the particles of gas depends only on the
temperature. It is the same for all gases at a given temperature, irrespective of the
masses of their particles/molecules.
The internal energy of a collection of particles in an ideal gas is defined as the sum
of all their kinetic energies. In a solid, liquid (or non-ideal gas), however, there are
forces between the particles - and so they also have potential energy. Thus the
internal energy of a solid, liquid or non-ideal gas is defined as the sum of the kinetic
and potential energies of the particles.
Redshift and Cosmology
Distant galaxies etc emit electromagnetic radiation that is of greater wavelength
(lower frequency) than would be observed from the same elements on earth. i.e.
Patterns of spectral lines are shifted towards the red end of the visible spectrum.
This phenomenon is known as redshift.
Redshift is reminiscent of the Doppler Effect (see SUR notes) - but the explanation is
somewhat different. In the case of redshift the wavelength has increased because
the whole universe has expanded since the waves were emitted by the distant
galaxy - and the waves have expanded along with it. For this reason the term
cosmological redshift is often used to distinguish this large-scale phenomenon
from the ordinary Doppler effect that would be observed if the light from a receding
planet or local star was examined.
Despite the above, it is possible to used the same formula that we used for the
Doppler effect in SUR .....
z =  = f / f = v / c
..... where v is the velocity with which the distant galaxy etc moving away from
us, c is the velocity of light, and  and f / f are the changes in wavelength
and frequency respectively, expressed as fractions. This fraction is also the
numerical value of z (the redshift).
Edwin Hubble (1889 - 1953) observed that the velocity of recession of galaxies
(measured using redshift) is proportional to their distance from us. Thus a graph like
the one below can be drawn .....
The velocity v is normally measured in kms-1 and the distance d is expressed in
megaparsecs (Mpc). 1 Mpc = 106 parsecs = 106 pc. 1 pc is the distance at which the
earth-sun distance would subtend an angle of 1 second of arc (1/3600 o), as shown
below .....
Thus 1 pc = 3.09 x 1016 m = 3.26 light years. (1 light year is the distance travelled by
light in one year.)
From the graph above ..... redshift z  d
And, provided v << c ..... z  v
Therefore ..... v  d
Or ..... v = Hod
(Hubble's Law)
..... where Ho is a constant called the Hubble constant. Ho is the gradient of a graph
of v against d. It is difficult to measure the Hubble constant, but the currently
accepted value lies in the range 50 - 100 kms-1Mpc-1. (The large uncertainty in Ho is
due to the large uncertainty in the measured distances of galaxies.)
Substituting v = zc into the Hubble's Law equation above, we obtain .....
z = Hod / c
Hubble's Law is related to the expansion of the universe - which (it is thought)
started at time zero with a "big bang", when all the matter/energy in the universe
was concentrated at a single point. Since then the galaxies etc have been moving
further and further apart, rather like raisins in dough, which is expanding ("rising") in
an oven .....
If we lived on another "galaxy" (raisin) it would also appear as though all the other
"galaxies" were moving away from us - and the further the "galaxies" are from us
the faster they appear to be moving - which is Hubble's Law.
There is an important link between the value of the Hubble constant Ho and the age
of the universe .....
Suppose a galaxy has been moving at a speed v for a time t since the Big Bang. (i.e.
t is the age of the universe.) The distance d it has travelled is given by .....
d = vt
But ..... d = v / Ho
(Hubble's Law)
Therefore ..... v / Ho = vt
Therefore ..... t = age of universe = 1 / Ho
(N.B. In order to calculate the age of the universe in seconds, the Hubble constant
must be expressed in ms-1m-1 (i.e. s-1) rather than kms-1Mpc-1.)
If the above calculation is done, the age of the universe turns out to be between 10
billion years (for Ho = 100 kms-1Mpc-1) and 20 billion years (for Ho = 50 kms-1Mpc-1).
Clearly the universe cannot be younger than the objects in it - and some galaxies
have been reliably measured to be 15-17 billion years old - which puts the lower
value for the age of the universe in doubt.
There is also uncertainty and controversy about the ultimate fate of the universe.
This depends on how strong the forces of gravity are which oppose the expansion of
the universe. (The strength of the gravitational forces depends on the average
density of matter in the universe.) If the density is high the expanding universe will be
pulled back by gravity into a "big crunch". This scenario is also referred to as a
closed universe. If the density is low the universe will go on expanding for ever - an
open universe. There is also the possibility of a critical universe, poised between
the previous two alternatives. These scenarios are illustrated below, together with
another possibility - the oscillating universe .....
The "shapes" of these universes are also different. A closed universe would be
finite (but with no boundary or "edge"). That is to say, space would curve back on
itself - rather in the same way that the surface of a sphere is curves back on itself,
with no edge. The critical and open universes would be infinite.
It is thought that the critical density c of matter that would produce a critical
universe is about 10-26 kgm-3. Suppose  is the actual density of our universe. We
can then introduce a quantity  which is given by .....
 =  / c
<1
=1
>1
open universe
critical universe
closed universe
Measured values of the average density  of the universe based on observations of
visible matter (stars, galaxies etc which are hot and giving out light) make  as low
as 0.1 - pointing towards to an open universe. However, our knowledge about the
rate of expansion of the universe suggests that the universe is actually very close to
being critical. It has therefore been proposed that there is a vast amount of unseen
or dark matter in the universe which would raise the value of  to 1. This dark
matter may be partly in the form of planets, dim stars or any other relatively cool
material which can't be seen from a distance. However, there is reason to believe
that most of it must be non-baryonic - e.g. neutrinos or WIMPS (Weakly Interacting
Massive Particles).