Physics notes, will add as nite goes on

[Sean]

for all those physics people who are now panicking like i am just putting up a few lil notes that may help refresh memory. if nothing else posting it helps me =)

*light behaves in a stream of photons which is illistrated by the photoelectric effect
* when a piece of metal is illuminated by electromagnetic radiation the energy is absorbed by electrons near the surface of the metal which can liberate them from their 'bound state'
*while increasing the intensity has no effect on the kinetic energy of the photoelectrons leaving the metal, however increasing the frequency does increase the kinetic energy that the photoelectrons leave with
*for each metal there is a 'threshold frequency' that the radiation iluminating hte metal must be greater then in order to excite the electrons enough for them to leave the metal. -- "a certain amount of energy had to be imparted to an electron on the metal surface in order to liberate it" which is known as that metal's 'work function'
*Bohr created a new model of the atom, after finding the rutherford model inadequet(sp?)
*bohr says that electrons orbit the nucleus at a certain radius. when it is constantly in this orbit it does not lose energy. however if the electron absorbs energy it is 'excited'!! to a higher orbit with a greater radius. after a short time in the excited orbit it returns to the lower orbit emitting a photon in the proccess
*the equation of the energy at each energy level is given by the equation of (Z^2/n^2) * (-13.6eV) where Z is the protons in the atom and n is the energy level
* the negative in the -13.6ev is due to the fact that electrons are bound to the atom and thus energy must be applied to cause it to leave: in other words bring it to a higher energy level
*when the electron drops to a lower energy level it relases a photon of energy that is the difference betwean teh 2 energy levels. the wave length of that photon is equal to (h*c)/(Ei-Ef)
*when you are finding what photons are expelled while the electron goes from one higher energy level to a lower energy level: it can give off a photon for every energy level it passes on its way to its destination. eg- in a photon going from N=4 to N=1 it can expell 3 photons. one for going from N4-->N3, one from N4-->N2 and one from N4-->N1

more later

[Mike]

Horrors of Euro forum coming back............

[Sean]

*de Broglie noticed that all units move in a wave. he found the wavelength to equal h/p (note h here is in J*s not in eV*fs) where for things wiht mass p (momentum) is equal to m*V
**a helpful note for this when using electrons or protons is **the use of Pe=Ke or qV=0.5mv^2 or
**Velocity =(2*m*q * Volts)^.5
*shutup mike
*conversion betwean mass and atomic mass unit (u) is 1u=1.6605*10^-27kg
*when comparing the mass of a nucleus and the sum of the masses of all the particles in it the sum of all the masses is greater then the mass of hte nucleus. this is due to the difference in mass was converted to energy when the atom was formed. this amount of energy is also the amount htat is needed to break the atom into its protons and neutrons. this energy is the "binding energy"
*1u = 931.6 MeV or 9.31*10^8 eV

[ean]

forget physics, lets play poker.

[Kob0724]

Well i'll tell you this right away. Your first note it wrong. Maybe for all intensive purposes (like the ones you use in your physics class) its ok to say that. But actually, light acts a wave as well. This was demonstarted magnificenly in feynmans famous experiment which he claims all of quatum physics can be derived from. See even if you were to slow it down so that the gun you were shooting the photons out of in your photoeletic experiment so that you were firing 1 individual photon each time, you would still get the interference patter. You might ask youself how this is possible. The photon has to go through one of the slits. If there are no other photons to effect its path of travel how come will still get the interference patter. Because light also acts as a wave my friend. Think of it like this: When the photon is fired out of the gun there are an infinite amount of paths it can take to get to any one point on the metal. Some of these paths have it going through the left slit, some going through the right, some going through both, some not going through either, and in some of the paths the photon goes out to the andromeda galaxy and come back. The point is all of these theoretical paths cancel out, except one. The remaining one is what you actuall observe on the metal. You can assign a probablity to each path that it will be the one that gets choosen. All of these probabilities taken into account together represent the photons "probability wave." For further questions contact me (Kurtis) or read the book "The Elegant Universe." I have taken no physics classes, just read on my own time. w00t.

[VanBen]

That is a beautiful book, kurtis

[Sean]

you are right in that light acts as both a wave and a particle (light duality) but that as nothing to do with the presance of photons. each photon that is emitted by a given source has a frequency and thus wavelength. so what you said is true, but irrelivant for what the purpose of that note is (pluss the fact that it has wave features is expressed in the final wording that it is illistrated in the photoelectric effect)