Sensei

I don't have iPhone, but Samsung. Typically I am simply connecting USB cable (it's part of phone package) between them and phone asks in which mode it should run. One of modes is to work as disk storage. (Other mode is f.e. use phone as modem to connect computer to Internet, people use it very often in hospitals. Needs phone & laptop, but pricey as hell. People usually disable images & flash in browser to not bankrupt) After that computer see my phone as yet another disk with the first free letter. f.e. H:\ You can then use Windows explorer to visit that location, and copy paste files like you would do with any other remote flash USB memory stick f.e. Another option is using FTP and/or HTTP. If your computer has public IP, you can set up FTP server, HTTP server, and use phone web browser you visit your own computer's server. Better to have wifi local connection than use phone Internet transmission which is pricey. If you don't have public IP, you can still use external FTP/HTTP servers though. But that would be even slower (have to upload file to external server, and then download it using phone browser).

Exactly. Anyway it is visible from equations: [latex]E = \frac{h*c}{\lambda} = \frac{6.62607*10^{-34} * 299792458}{650*10^{-9}} = 3.056*10^{-19} J[/latex] [latex]E = \frac{h*c}{\lambda} = \frac{6.62607*10^{-34} * 299792458}{525*10^{-9}} = 3.784*10^{-19} J[/latex] [latex]E = \frac{h*c}{\lambda} = \frac{6.62607*10^{-34} * 299792458}{405*10^{-9}} = 4.9*10^{-19} J[/latex] quantity of red photons in [latex]1 W = 1 \frac{J}{s}[/latex] is [latex]\frac{1 J}{3.056*10^{-19} J} = 3.27*10^{18}[/latex] photons per second. quantity of green photons in [latex]1 W = 1 \frac{J}{s}[/latex] is [latex]\frac{1 J}{3.784*10^{-19} J} = 2.64*10^{18}[/latex] photons per second. quantity of blue photons in [latex]1 W = 1 \frac{J}{s}[/latex] is [latex]\frac{1 J}{4.9*10^{-19} J} = 2.04*10^{18}[/latex] photons per second.

In most cases it's correct. There is yet another explanation: white light is falling on object, then reflected photons are reacting us, but other also reflected photons with different frequencies, or at different angles, are not reaching us (instead of absorption). It's called Iridescence. http://en.wikipedia.org/wiki/Iridescence Also instead of absorption there can happen refraction. Majority of black body objects are emitting light at frequencies invisible to human eye. For instance infra red, or microwave. You can see them emitting light (especially when they're really hot) using IR camera. This can be used to detect fire, before flame really appeared. Photon with frequency f, has energy E=h*f f = c/wavelength and wavelength = c/frequency so wavelength = h*c/E My red photon laser has wavelength 650 nm, green photon laser has wavelength 525 nm (24% more energy than red), blue/violet photon laser has wavelength 405 nm (30% more energy than green, and 60% more than red). So it's not just matter of "color", it's matter of energy. Energy can be used to heat substances. Red photons will heat substances to lower temperature, than green, than blue (if we send them in the same quantity) (assuming they will be all absorbed). Photons with certain frequencies (energies) trigger (or not) chemical reactions.

But it wouldn't fit to idea of annihilation of electron and positron that happens later. To do so, there is needed that they're more or less point particles, that meet together and finally annihilate. Until they don't annihilate, they're existing as e+ and e-. If you can keep them separate (like in magnetic trap), they won't annihilate. When electron accelerated to relativistic velocity collides with positron, there are created other particles. "There are also many examples of conversion of relativistic kinetic energy into rest energy. In 1974, SLAC National Accelerator Laboratory accelerated electrons and positrons up to relativistic velocities, so that their relativistic energy \gamma mc^{2} (i.e. the sum of their rest energy and kinetic energy) is significantly increased to about 1500 MeV each. When those particles collide, other particles such as the J/? meson of rest energy of about 3000 MeV were produced.[30] Much higher energies were employed at the Large ElectronPositron Collider in 1989, where electrons and positrons were accelerated up to 45 GeV each, in order to produce W and Z bosons of rest energies between 80 and 91 GeV. Later, the energies were considerably increased to 200 GeV to generate pairs of W bosons" Quote from http://en.wikipedia.org/wiki/Tests_of_relativistic_energy_and_momentum It's result of cascade ionization I mentioned before. Similar ionization happens when charged particle is passing through liquid Hydrogen in f.e. Bubble Chamber. But that's not the point. If electron would be as "wide" as orbital of let's say Hydrogen, it's path in chamber should looks like zig-zag constantly changing directions after hitting nucleus of medium. Alpha particle has a few femtometers and is occasionally hitting medium nucleus and changing directions, like here: http://www.nuffieldfoundation.org/practical-physics/alpha-particle-tracks-including-collision-helium-nucleus Free electron is not limited to any orbital... One might ask, which atom orbital, and with what quantity of electrons. If you have f.e. Gold nucleus with 79 protons, and 1 electron, ionization energy is approximately 13.6*Z^2 = 13.6*79^2 = 84.8776 keV In other words you must send photon with ~85 keV to remove the last electron from ion 79Au78+ Such molecule is "Hydrogen-like", all charge concentrated in nucleus (+79e), and single electron outside (-1e). Simply: finding where was our point particle.. You completely forgot that whole thread is your own hypothesis/speculation that electron is point particle, and now you're defending opposite...

I thought you wanted free discussion ("Your thoughts please?"), but unfortunately it seems I was wrong..

See the way electrons are created in pair production: y + y -> e+ + e- Each photon with E >= 510999 eV, which is wavelength equal to Compton wavelength = 2.426*10^-12 m = 2.426 pico meters. or with single photon and nucleus: y -> e+ + e- E = 1.022 MeV, wavelength half of Compton wavelength. After pair production we will have electron with half energy of such photon, and positron with another half of that initial energy. Electron particle "bigger" than this wavelength, wouldn't make sense to me.. Another argument for small size of electron is positronium: http://en.wikipedia.org/wiki/Positronium Electron and positron are orbiting together making exotic atom. You're thinking too much (and overestimating) orbitals. While you should concentrate on free electrons, not bound to any atom IMHO. See electron's trace which it's making in Cloud Chamber: Electron had to be in the center of that trace. Electron accelerated to significant speed of light, with tremendous amount of kinetic energy (see decay energy calculations to find out it's possible maximum kinetic energy for given unstable isotope (that nucleus had mass-energy concentrated in couple femtometers *)). Passing through medium, it gave part of its kinetic energy to medium, ionized it, and accelerated atom (or its electrons) of medium hit other medium atom - cascade effect happened (and repeat it along path for billions of billions of medium particles). After giving all its kinetic energy, it's stopping moving. What happened at quantum, picometer or femtometer scale, is therefor visible in our scale by naked eye. It would be worth to record it in slow motion 1000+ FPS camera, to see how trace is growing with time. *) According to http://en.wikipedia.org/wiki/Femtometre "radius of a proton is approximately 0.84 - 0.87 femtometres" Now imagine proton (in nucleus) decaying via beta decay+, to positron p+ -> n0 + e+ + Ve or neutron decaying via beta decay-, to electron n0 -> p+ + e- + Ve IMHO electron radius should be even smaller than nucleus that decayed (and while before decay was "containing" our electron/positron).

I have personally performed test with 2 meter long Aluminum straight pipe, 1 cm diameter. And couple (quantity does matter!) neodymium cylindrical magnets, 8 mm diameter, 4 mm height each. With 5-6 such magnets, I was receiving steady 12-13 seconds time spend on moving to the other side of pipe (pipe vertically of course). That's 19-20 times longer than free fall. Free fall 2 meters takes ~0.64 seconds normally (and that's delay when we pass anything but these magnets through my pipe). I will make video when will buy new Full HD camera.. Results (today performed again to be sure): 1 magnet: ~16 seconds 2 magnets: ~21 seconds 5 magnets: ~12 seconds

Have you seen such home experiment?

No. I don't think so, if you meant "nuclear binding energy". You should calculate decay energy of the all isotopes of the all mentioned elements for the all possible case of decay (or at least for proton, neutron and alpha decay). For stable isotope, decay energy is negative value. And that's energy you need to spend to separate nucleus to smaller parts. Then you can sort them in right order. f.e. binding energy for isotope Carbon-12 is different than for isotope Carbon-13. Either one are stable. But it's awful lot of work. Tin for example has 10 stable isotopes http://en.wikipedia.org/wiki/Isotopes_of_tin You should ask your teacher whether he meant "average nuclear binding energy per nucleon". If so, you should calculate energy of the all free protons and free neutrons in isotope with Z protons and A-Z neutrons. Subtract from it nucleus energy (isotope mass * c^2 - mass-energy of the all electrons). (after dividing it by A, you have average binding energy per nucleon).

The same can be said about any moving macroscopic object like f.e. airplane. If we know position of airplane, it must be on land, not flying anymore. If it's flying in air, its position is constantly changing, and while we're seeing it, it's not there anymore, we just see photons reflected from it after some time. And our recorded position is more or less not actual. You're talking about electron in electronic circuit. Double slit experiment for electrons is done in vacuum with free electrons, emitted by electron gun. Detection of electron means it has hit detector. And perhaps created photon, which has been seen, and counted. Electron slowed down, gave detector it's kinetic energy, or part of it. And it no longer poses that kinetic energy. Act of "observation", measurement, caused change in object we're measuring.

IMHO it's good to visualize where are electrons, where they go, and which path they will be flowing. Electric current I is going in opposite direction to flow of electrons (because XIX century scientists developed it up-side-down). Take for example Faraday's disc http://en.wikipedia.org/wiki/Homopolar_generator http://en.wikipedia.org/wiki/Faraday_wheel Close to metal wheel there is placed magnet (but not touching it), wheel is spinning, and electrons in metal are repelled by magnets. It happens regardless of whether circuit is closed or not. Electrons gather in part of wheel not affected by magnets. We can place there wire f.e. a few cm from outside of wheel, electrons will ionize air gap creating spark and will be going through wire, then where ever we want. It might be used during electrolysis, which needs large current, with low voltage. Four electrons, each with E = 1.23 eV will be able to split two water molecules, and produce 2 H2 and O2

Who said so.. ?

Nobody is saying about pattern. I just said "pattern is completely not needed". bool IsPrime( unsigned long long value ) { if( value < 2 ) return( false ); unsigned long long max = (unsigned long long) sqrt( (double) value ); for( unsigned long long i = 2; i <= max; i++ ) { if( ( value % i ) == 0 ) { return( false ); } } return( true ); } If prime would be random, chance to return true or false for given input value, would be 50/50. Instead there is returned true, if value is prime, or false, if it's not. If we would be interested about even numbers, it'd be also returning true or false, but inside there would be: return( ( value % 2 ) == 0 );

Nonsense. Primes don't fulfill definition of random f.e. http://www.macmillandictionary.com/us/thesaurus/american/random "chosen or happening without any particular method, pattern, or purpose" Probability of that 2,3,5,7,11,13 etc etc. is prime number is 100%. Pattern to predict next prime is completely not needed. You don't ask for next even number while examining f.e. 10. There is route to check whether some number is prime or not, by dividing it by all lower numbers. Definition of even is that it's dividable by 2. 10 is dividable by 2, so it's even. We're not interested about any lower, or higher numbers at all.

Where do you have source of sound at such frequencies? Do you have equipment generating such sound? Do you have equipment detecting such sound? If you do so, then start placing different materials between emitter and receiver, to see which one will be working.. Do you know how sound is generated by speaker? By moving membrane in and out, in and out, as many times as frequency you have in sound. If membrane would be moving 3 GHz for 1 cm it's 3*10^9 * 2 * 1 cm = 60 mln m/s = 20% speed of light. For 1 mm distance movement, it's 2% speed of light. At 16 kHz membrane moving 1 cm in & out, is reaching nearly speed of sound in air.

The best software you can have, is your own software. You can change it any time you need and want.. And it's free..

This theory should be easy to check.. Place piece of graphite parallel and then perpendicular to applied current. If you're right, results should differ. ps. What I coincidence. I was just two hours ago checking what magnetic field lines will be created by piece of graphite through which we are passing 2 A current.

Not exactly. It's muon's neutrino. or muon's anti-neutrino. Depending on charge. There are three types of neutrinos discovered so far: electron neutrino, or anti-neutrino. muon neutrino, or anti-neutrino and tau neutrino, or anti-neutrino. You won't detect muon's or tau's neutrino using neutrino detector designed for electron's neutrino. f.e. Chlorine-37 based neutrino detector will react only with electron's neutrinos with E >= 0.813 MeV. If neutrino has less energy (f.e. produced in ordinary fusion of Hydrogen) neutrino's detector won't be triggered.

Bond-dissociation energy of Hydrogen is completely different thing. 4.52 eV per molecule, and 436 kJ/mol. It's three times lower than ionization energy. http://en.wikipedia.org/wiki/Bond-dissociation_energy If you will dissociate molecule: H2 -> H+ + H- You will have one proton with 2 electrons, and one proton with 0 electrons. 12.75 eV excites electron in Hydrogen from n=1 to n=4 as has been showed in previous post. and 10.2 eV excites completely different electron in completely different Hydrogen from n=1 to n=2. After excitation, it's releasing photon, and goes back to ground state. Then it can absorb another one.

Where did you get that 50%?? Made up by yourself... Nobody is saying about 50% probability in case of any radioactive isotope... Are you familiar with calculation of decay rate using half-life? [latex]I = I_0 * 2^{-\frac{t}{t_1/2}}[/latex] I0 - initial quantity of radioactive isotope (or unstable particles) I - current quantity of radioactive isotope t - current time t1/2 - half life time

I am repeating: Electron absorbed photon with energy equal to [latex]E = \frac{13.6 eV}{1^2} - \frac{13.6 eV}{4^2} = \frac{13.6 eV}{1} - \frac{13.6 eV}{16} = 13.6 eV - 0.85 eV = 12.75 eV[/latex] Electron absorbed photon with energy equal to [latex]E = \frac{13.6 eV}{1^2} - \frac{13.6 eV}{2^2} = \frac{13.6 eV}{1} - \frac{13.6 eV}{4} = 13.6 eV - 3.4 eV = 10.2 eV[/latex] Send photon with energy E = 13.6 eV and you will permanently ionize Hydrogen atom. It means we will have free proton and free electron not bound together. 13.6 eV * 1.602*10^-19 = 2.17872E-018 J * 6.022141e23 mol^-1 = 1312056 J/mol = 1312 kJ/mol And such value you can see on element's ionization table f.e. here: http://en.wikipedia.org/wiki/Ionization_energies_of_the_elements_(data_page) That's energy needed to make plasma from the all Hydrogens in 1 mole of particles. Once particle is proceeding other particle, thus photon emitted will be blue shifted, and other time it's receding other particle, thus photon emitted will be red shifted and won't be able to be absorbed by that 2nd particle. It's relativistic Doppler shift. This is what makes spectrum lines wider. The faster particles are moving the wider spectral line. If you send beam of photons with E = 12.75 eV to Hydrogen, you will receive photons with E=10.2 eV and E=2.55 eV If you send beam of photons with E = 13.6 eV to Hydrogen, you will receive photons with E=10.2 eV, E=2.55 eV, E=1.88 eV, and many many others that you can calculate by yourself using above equation. See this screen-shot what energy photons you will receive: Nope. Ionized Hydrogen atom has no electrons. There is free proton, and somewhere else there is free electron. Where did you get such information that atom in ground state will be emitting photons? Nucleus can emit gamma photons, but that's completely different than what we're speaking about.

Do you even know how these results are obtained in lab? By ionization of gas in tube by high voltage, then passing it through prism, and then after zooming to screen. If you send photon with specific energy, you will excite atom(s) to specific maximum level. Excited electrons in atom will emit photons. Some photons might be absorbed by other atoms, and new photons emitted. Excited electron might not be able to return to initial level, because it's already taken by other electron (after emitting other photon with different energy). They will swap their "locations". Cause is high voltage. You can build your own spectral lines tube using f.e. glass test tube and fill it by Hydrogen, Oxygen from electrolysis. And connect it to Cockcroft-Walton generator. And see it on your own eyes. Search on eBay for "discharge tube" or "spectral line tube" f.e. http://www.ebay.com/itm/New-Spectrum-Discharge-Tube-x-2-Spectrum-Analysis-Gas-Spectrum-Tube-set-of-2-/171435386822 $45 for two tubes is good price IMHO.

??? Electron goes from n=1 to n=2,3,4,5.... infinity (theoretically) when it absorbs energy, usually from photon. And from higher n to lower, when it's releasing photon. Energy of photon (E=h*f) defines to which level electron will be able to go. f.e. [latex]E = \frac{13.6 eV}{n_1^2} - \frac{13.6 eV}{n_2^2}[/latex] For n1=1 and n2=2 [latex]E = \frac{13.6 eV}{1} - \frac{13.6 eV}{4} = 10.2 eV[/latex] For n1=2 and n2=3 [latex]E = \frac{13.6 eV}{4} - \frac{13.6 eV}{9} = 1.88889 eV[/latex] This is actually in visible spectrum, wavelength = 656 nm Red spectral line in spectrum of Hydrogen. For n1=2 and n2=4 [latex]E = \frac{13.6 eV}{4} - \frac{13.6 eV}{16} = 2.55 eV[/latex] This is in visible spectrum, wavelength = 486 nm

There is theoretical work that limits distance cosmic rays with relativistic speeds can travel. It's based on theory that when relativistic particle will collide with background radiation photon, there might be created pion particle because of large relativistic Doppler blue shifting. http://en.wikipedia.org/wiki/Greisen%E2%80%93Zatsepin%E2%80%93Kuzmin_limit