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Astro Final
| Question | Answer |
|---|---|
| What does X stand for in atmospheric extinction? | sec(z) or the airmass |
| Atmospheric Extinction | To know the quality of the atmosphere on any night, you need to measure the atmospheric extinction coefficient, observe the same star every few min/hrs so you catch it at different air masses. Measure the magnitude of the star at each air mass. plot |
| Typical dust grains preferentially | scatter blue light (shorter wavelengths) but allow red light (longer wavelengths) to pass through more easily, thus B increases (gets fainter) relative to V |
| When B increases (gets fainter) relative to V: | So there is more red light than blue light, so the object appears redder than it should. (B-V increases) |
| Amount of scattering relies on | Wavelength |
| Why is the sky blue? | Reddening is caused by red light preferentially making it through the extincting material. The blue light is preferentially scattered |
| B-V | (mb-mv) observed colour index. Affected by extinction |
| (B-V)0 | (Mb-Mv) Intrinsic colour index. related to temp and unaffected by extinction |
| E(B-V)= | Ab-Av The color excess |
| So, if you can measure the observed colour (always possible) and you know the intrinsic colour (if you know the temperature) then | You can calculate how many magnitudes if visual extinction the radiation had to have passed through to get reddened by the observed amount |
| Planck Curve or Black Body Spectrum | Is called Distribution of energy. All sufficiently dense objects radiate. |
| A perfect black body is | one that absorbs all energy incident upon it, comes to some equilibrium temperature and re-emits an amount of radiation dependent on that temperature. No reflection |
| Bv(T) is | the intensity as a function of frequency and temperature per unit area of the emitting object |
| Red emission appears | reddish in colour when the red emission is brighter than blue emission |
| Blue emission appears | blueish in colour when the blue emission is brighter than red |
| Every temperature has | a unique ratio of red to blue light. For a BB, a specific (measured) ratio of red/blue light means that the BB has a specific temperature. We can tell something about temp of a distant star from its colour |
| To actually determine the temperature from photometric observations, we want to take the ratio of fluxes that have been measured in two filters | Essentially integrating the Planck function over the small wavelength/frequency ranges that the filters span. So we measure a wavelength integrated flux in a particular filter |
| Planck function- if two stars have the same size, | the hotter star has a higher luminosity |
| Planck function-if two stars have the same temperature | the larger star has a a higher luminosity |
| Total energy (luminosity) emitted from the BB is obtained by | integrating over the entire surface of the emitting sphere: L=4*PI*R^2*F |
| Wien's law | We can get a precise measure of its temperature by measuring the wavelength of the peak of the BB curve. Notice the peak of the BB curve shifts to higher frequency (shorter wavelength) as T increases |
| How can you use Wien's law with B(T) | take the derivative with respect to wavelength and set it equal to zero and find at what wavelength does the slope equal 0 |
| Temps we get from Wien's law and Stefan-Boltzman law are | effective temperatures. |
| Te is | The temperature of a BB which radiates with the same total flux as the object (since no object is really a BB) |
| Wien's Law is | possibly a little more accurate than the B-V method since you're looking at more global properties rather than individual bands which could be affected by absorption |
| Spectral lines | Low density (gaseous objects also emit light). NO TWO SPECIES ARE ALIKE. Both atoms and molecules emit light at a small number of frequencies, and no others. These frequencies are particular to the emitting species |
| One of the great successes of modern astronomy is the ability to identify atoms and molecules in astronomical objects | without ever having to be in direct contact with them. Thus, is possible to compare the chemical composition of say, the sun, to other stars. |
| Spectral lines and neutral atoms | Radiation emitted by neutral atoms and molecules does not occur at any random frequency (or wavelength) |
| The pattern of lines emitted by an atom | (the frequency/wavelength of light emitted by an atom) is directly related to the structure of the atom |
| The hydrogen atom | consists of one electron orbiting around the nucleus composed of a single proton like a satellite orbiting around the earth |
| The hydrogen atom and energy | The distance between the orbiting electron and the nucleus is proportional to the energy of the orbit. The various orbits of the electron are called energy levels. H has an infinite number of states but they get scrunched together |
| The hydrogen atom and quantum | The electron is only allowed to orbit in very specific energy levels. We say that the electron's energy levels are quantized. One of the fundamental rules of atomic physics |
| The hydrogen atom and energy levels | In general, the electron will be orbiting in level 1 (ground state) where the electron always "wants" to be. But the electron could be orbiting the nucleus in any one of an infinite number of discrete energy levels |
| When a photon strikes a hydrogen atom, one of three things can happen | ionization, de-excitation (emission) or the photon and electron will ignore one another |
| Ionization | The photon will have so much energy that it knocks the electron off of the atom, leaving an ionized hydrogen atom behind |
| De-excitation producing spontaneous emission of a photon | If the photon's energy exactly matches the energy difference between two levels in the hydrogen atom, the photon will be absorbed by the atom. |
| Excitation which produces absorption of a photon | In the process of de-excitation where the energy of the photon matches exactly the energy difference. The atom acquires the energy of the photon and the electron "jumps" to a higher energy state. |
| After excitation in the de-excitation process | after a very short period of time: the electron "wants" to get back to the ground state so it spontaneously "jumps" back to some allowed lower energy level. e emits a photon with an energy exactly =to the E diff between the two levels |
| Excitation | absorbs a photon of energy E and causes electron to jump to a higher energy level. The energy difference between the 2 levels is also change in E |
| De-excitation | Produces spontaneous emission of a photon of energy delta E and causes electron to jump to a lower energy level. The energy difference between the 2 levels is also delta E |
| Incoming and outgoing photons... | do not necessarily have the same direction!!! |
| If the incoming photon does not have the right amount of energy to exactly match the difference in the atom's energy levels | The atom and photon will ignore each other completely. The photon will continue on as if the atom wasn't even there |
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