E.2.7 to E.2.11: Astrophysics

E.2.7 Explain how atomic spectra may be used to deduce chemical and physical data for stars.


The radiation from stars is not a perfectly continuous spectrum-- some wavelengths are missing. The missing wavelengths correspond to a number of elements. The absorption takes place in the outer layers of the star therefore this means that we have a way of telling which elements are in the star (or at least in its outer layers).

A star that is moving relative to Earth will show a Doppler shift in spectrum.

Red shift: light from stars that are receding
Blue shift: light from stars which are approaching.



E.2.8 Describe the overall classification system of spectral classes.


Different stars give out different spectra of light which allows us to classify stars by their spectral class. Stars that emit the same type of spectrum are allocated to the same spectral class.

There are several main spectral classes in order of decreasing surface temperature.

O: Oh

B: Be

A: A

F: Fine

G: Guy

K: Kiss

M: Me

E.2.9 Describe the different types of star.


BINARY STARS:

• Some stars like our Sun exist by themselves, but many have a partner
• Binary stars rotate around their own common centre of mass
• By analysing the orbital period and separation, the mass of each star in the binary system can be found

RED GIANT STARS: 

• Large in size
• Red in colour
• Comparatively cool
• Later possible stages for a star
• Source of energy: Fusion of some elements other than hydrogen
• Red supergiants are even larger

WHITE DWARF STARS:

• Small in size
• White in colour
• White therefore comparatively hot
• One of the final stages for some stars
• Fusion is no longer taking place - white dwarf is just a hot remnant that is cooling down
• Eventually it will cease to give out light when it becomes cold enough
• After ceasing to give out light (cold) it is called a brown dwarf 

CEPHEID VARIABLES:

• They are stars that are a little unstable.
• Observed to have a regular variation in brightness and (therefore) luminosity
• Aforementioned variation due to oscillation in the size of the star
• Rare but useful because there is a link between period of brightness variation and average luminosity
• Astronomers can therefore use them to help calculate the distances to some galaxies

E.2.10 Discuss the characteristics of spectroscopic and eclipsing binary stars.


VISUAL BINARIES
Binary stars (e.g. Sirius A) that can be seen with the naked eye of with a telescope are called visual binaries, when they are further away from us or closer together, resolution become difficult.

SPECTROSCOPIC BINARY STARS
In some cases, stellar spectra can be used to deduce the presence of two stars - these are called spectroscopic binary stars. As stars move around their common centre of mass, one star will be approaching whilst the other is receding.

The diagram above shows a spectroscopic binary system. In the right hand diagram, Star A approaches and Star B recedes from our line of sight. Therefore the absorption lines of A are blue shifted  and the absorption lines of B are red shifted (moving away so longer wavelength, therefore red). In the left hand diagram, because the motion of the stars relative to our line of sight is opposite, the shift is reversed.

ECLIPSING BINARY STARS
Show a periodic variation in the brightness of light emitted from the star system. This occurs because during their rotation, one star periodically obscures, or eclipses, the other.

The diagram above shows an eclipsing binary system.
Position-
1 and 3: light is reaching us at a maximum, because it is arriving directly from both stars
2 and 4: reduction in brightness as the stars are eclipsing each other.

E.2.11 Identify the general regions of star types on a Hertzsprung-Russell (HR) diagram.


Discovery by Hertzsprung and Russell in 1910: For most stars, there is a relationship between surface temperature and luminosity. 

Dots on diagram below represent stars, scales are not linear.

Temperature scale: Runs backwards, high temperatures on the left.
Absolute magnitude: The apparent magnitude it would have if it were observed from a distance of 10 parsecs. Absolute magnitudes are much more negative than the apparent magnitudes of the stars.

• 90% of stars fall into the diagonal band known as the main sequence. It can be shown using the Stefan-Boltzmann Law that stars increase in size as we move up the main sequence.
• Lower right: The coolest stars, reddish in colour.
• In the middle: (further towards the left than lower right) hotter, more luminous stars that are yellow and white.
• Lower left: more luminous blue stars.

Mass of star increases moving up the main sequence so the gravitational pressure increases with mass. Therefore, to maintain equilibrium, fusion reactions in the core must generate a greater radiation pressure. The star has to burn at a higher temperature, giving it a greater luminosity.


9% of stars = red giants and supergiants. From Stefan-Boltzmann Law we can see that their high luminosity and low temperature means that they must have a very large area - they are therefore giants. 


WHITE DWARFS are very hot but not luminous, therefore they are much smaller than their counterparts on the main sequence.

The cepheids, congregate in a great band of instability that appears between the main sequence and the red giants.

Images and diagrams from: Heinemenn HL Physics Course Companion, IB Physics study guide by Tim Kirk

E2 Stellar Radiation

E.2.1 State that fusion is the main energy source for stars.

Fusion is the main energy source for stars.

E.2.2. Explain that in a stable star (for example, the Sun), there is an equilibrium between radiation pressure and gravitational pressure.

A stable star is a star in which there is an equilibrium between the radiation pressure and the gravitational pressure. The reason why the powerful reactions in the Sun have not forced away the outer layers of the Sun is because of this balance between the outward pressure and inward gravitational force.

E.2.3. Define the luminosity of a star.

Luminosity is the total power radiated by a star, it is measured in Watts (W). It depends on both the surface temperature of the star and its radius or surface area. If the radius of the two stars is the same, the one with the higher temperature will have the greater luminosity. If the temeperature of the stars is the same, the one with the larger radius will have the great luminosity.

E.2.4 Define apparent brightness and state how it's measured.

Apparent brightness: The power received per unit area. The SI units are Wm^-2.

E.2.5 Apply the Stefan-Boltzmann Law to compare the luminosities of different stars

The radiation from a perfect emitter is known as black body radiation. The graph below shows a spectrum of radiation from black-body emitters at different temperatures. A hot object emits radiation across a broad range and there is a peak in intensity at a particular wavelength. For a hotter body, the peak is at a higher intensity and shorter wavelength.

The peak wavelength (at which the maximum amount of energy is radiated) is related to the surface temperature by Wien's displacement law.

E.2.6 State Wien's (displacement) law and apply it to explain the connection between the colour and temperature of stars.

Questions 5 & 6

5.

Astrophysics

3) Make a pneumonic for the order of the planets:

Mercury

Venus

Earth

Mars

Jupiter

Saturn

Uranus

Neptune

My Very Educated Monkey Just Sat Uncomfortably Naked

4) Compare sizes of planets with circles

http://www.google.co.th/imglanding?q=planet%20sizes&imgurl=http://hyperphysics.phy-astr.gsu.edu/hbase/solar/imgsol/psize.gif&imgrefurl=http://hyperphysics.phy-astr.gsu.edu/hbase/solar/sizes.html&usg=__KzUp-rOcpR2Kx8pWWukoJPWSTJI=&h=335&w=342&sz=5&hl=th&zoom=1&um=1&itbs=1&tbnid=X0xyO42EuscbNM:&tbnh=118&tbnw=120&prev=/images%3Fq%3Dplanet%2Bsizes%26um%3D1%26hl%3Dth%26sa%3DN%26tbs%3Disch:1&ei=8g9iTa_MOo3NrQfIm4SxAQ&um=1&sa=N&tbs=isch:1&start=12#tbnid=X0xyO42EuscbNM&start=16

5) Define:

Asteroid: Small rocky body that drifts around the Solar System.

Meteoroid: An asteroid on a collision course with another planet.

Meteorite: Small meteors can be vaporized due to the friction with the atmosphere (‘shooting stars’) whereas larger ones can land on Earth. The bits that arrive are called meteroties.

Comet: Mixtures of rock and ice (a ‘dirty snowball’) in very elliptical around the Sun. Their ‘tails’ always point away from the Sun.

Stellar Cluster: Group of stars that are physically close to each other, created by the collapse of the same gas cloud.

Constellation: Patterns of stars.

Light year: The distance light travels in 1 year.

6) Compare distances between stars and between galaxies

(From Physics IB Study Guide by Tim Kirk) 

7) Describe the motion of constellations over

1 night

They appear to rotate around in one direction. In the Northern hemisphere, everything seems to rotate around the North Star.

1 year

(From Physics IB Study Guide by Tim Kirk) 

13.2.3 & 13.2.4 Nuclear Physics

13.2.3 Describe one piece of evidence for the existence of nuclear energy levels.

For example, alpha (α) particles produced by the decay of a nucleus have discrete energies; gamma‑ray (γ-ray) spectra are discrete. Students should appreciate that the nucleus, like the atom, is a quantum system and, as such, has discrete energy levels.



The energies of both alpha and gamma are discrete (quantized) and have only certain values. Since these discrete energies come from the nucleus, it must follow that the nucleus has discrete nuclear energy levels. The graph below shows this property of alpha and is adapted from Heinemann HL Physics book by Chris Hamper.

13.2.4 Describe β+ decay, including the existence of the neutrino.

Students should know that β energy spectra are continuous, and that the neutrino was postulated to account for these spectra.


When energy calculations were done with beta emissions, scientists observed that the beta particles had less kinetic energy than expected. This opposed the law of conservation of energy (and in fact, scientists were so puzzled by this problem that they started to question the law of conservation of energy itself). In 1930, Wolfgang Pauli postulated that there was a virtually undetectable particle that therefore carried away this missing kinetic energy and momentum. This particle was named a 'neutrino'.


A neutrino is electrically neutral, has a very very small mass and travels at the speed of light.


In positron decay, a proton within the nucleus decays into a neutron and a positron (antimatter version of an electron--its antiparticle, when the two collide, they annihilate each other) is emitted. The equation for is is shown below (extracted from IB Physics Study Guides by Tim Kirk):

 Because of neutrinos and anti-neutrinos, beta emission forms a continuous spectrum. This is shown below (graph adapted from Heinemann HL Physics book by Chris Hamper).

13.2.1 & 13.2.2 Nuclear Physics

13.2.1 Explain how the radii of nuclei may be estimated from charged particle scattering experiments.

In charged particle scattering experiments, the particle that is repelled straight back will initially be momentarily at rest before it changes direction to travel straight back. Rutherford thus proposed that this fact can be used to estimate the size of the nucleus. This is because at that point at which the particle is at rest, its kinetic energy will be exactly balanced by its electrical potential energy due to the repulsive electrostatic force. 

13.2.2. Describe how the masses of nuclei may be determined using a Bainbridge mass spectrometer.

A magnetic field is used to deflect moving ions of a substance. If the ions have the same charge they will have the same velocity, v. The radius of their circular path however depends on the mass of the ion. A larger mass ion will travel in a larger circle. Therefore, the masses of nuclei may be determined using a Bainbridge mass spectrometer.

19.4
λ= 3.19 x 10^-7
f = v / λ
= 3 x 10^8 / 3.19 x 10^-7
= 9.4 x 10^14
φ + KE = hf
hf= 6.24 x 10^-18
KE = 5.86 x 10^-18 J
φ= 3.78 x 10^-19


Vs=KE=Vs x q
5.86 x 10^-18= Vs x 1.6 x 10^-19
Vs = 36.6 V


19.5
f = v / λ
f= 3x10^8 / 5 x 10^-7
=6 x 10^14
φ + KE = hf
6.63 x 10^-24 x 6 x 10^14 = φ + 2.4 x 10^-15
φ= 1.878 x 10^-19 J


KE now= 9 x 10^-19 J
hf= 1.878 x 10^-19 + 9x 10^-19
= 1.0878 x 10^-18
f= 1.64 x 10^15


λ = v/f
=3 x10^8 / 1.64 x 10^15
=1.83 x 10^-7 m

Quantam Physics - Q1,2,3,4

1) a) Drawn on one note
b) i) Planck's constant = gradient of line
m = change in y / change in x
m= 1.6 x 10^-19 / (7.4 x 10^14 - 4.9 x 10^14)
m= 6.4 x 10^-34 Js
Close to Planck's constant 6.6 x 10 ^-34

ii) 

KEmax = hf -φ
y = mx + c
( - because its a negative y-intercept)
Finding φ at 1.6 x 10^-19 and 7.4 x 10 ^14 Hz,
6.6 x 10 ^-34 x 7.4 x 10 ^14 = φ + 1.6 x 10^-19
φ= 3.284 x 10^-19
= 3.2 x 10^-19


b) If the frequency is lower than threshold frequency then KEmax will be more than hf and so subtracting KEmax from hf would yield a negative value. The work function cannot be a negative value.


2) a) Changes that occur in the micrometer when...
i) The intensity of incident light is increased (but frequency remains the same)
An increase in intensity will mean that more electrons will be liberated. As Q= IT, more electrons will mean more charge per unit time (Q/T increases) and as I = Q/T, I (current) will also increase. The microammeter will therefore detect an increase in current.
ii) The effect of increasing frequency will depend as to whether the starting frequency is the threshold frequency or not. If it is and the frequency is increased above the threshold frequency, the maximum energy of the electrons will  depend on the frequency of the incident light and will be higher as the frequency gets higher. If it is below threshold frequency and is increased to a value still below the threshold frequency, no photoelectrons will be emitted the current is still zero.


b) i)
 ii) Ek = 1.9 x 1.6 x 10^-19 = 3.04 x 10^-19 J
540 nm = 540 x 10^-9 m
v= fλ
3 x 10^8 = 540 x 10^-9 x f
f= 5.56 x 10^14 Hz
φ = hf - KEmax
φ= (6.6 x 10 ^-34 x 5.56x10^14) - 3.04 x 10^-19
φ= 6.3 x 10^-20




3) a) The de Brogile wavelength of a particle is the wavelength of a wave that determines the probability of a particle's position.

b) 
5.0 kV= 5.0 x 10^3 V
5.0 x 10^3 x 1.6 x 10^-19= 8 x 10^-16 J
Ek = 1/2 x m x v^2
8 x 10^-16 = 1/2 x 9.11 x 10^-31 x v^2
v= 4.2 x 10^7 ms^-1
p=mv
=9.11x 10^-31 x 4.2 x 10^7
p=3.8 x 10^-23


λ = h / p
λ= 6.6 x 10 ^-34 / 3.8 x 10^-23
λ= 1.7 x 10^-11


4) a) V= 75 V
Ek = 75 x 1.6 x 10^-19 = 1.2 x 10 ^-17
Ek = 1/2 x m x v^2
1.2 x 10 ^-17= 1/2 x 9.11 x 10^-31 x v^2
v= 5.1 x 10^6 ms^-1
p=mv
p= 9.11x10^-31 x 5.1 x 10^6
p= 4.67 x 10 ^-24
λ = h / p
λ = 6.6 x 10 ^-34 / 4.67 x 10 ^-24
λ = 1.4 x 10^10 m


b) The general shape of the graph mirrors the shape of a diffraction pattern as can be easily discerned from the maxima and minima. This graph therefore confirms de Brogile's hypothesis that electrons have both wave and particle properties.