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Proton NMR From Wikipedia, the free encyclopedia Jump to: navigation, search Proton NMR (also Hydrogen-1 NMR, or 1H NMR) is the application of nuclear magnetic resonance in NMR spectroscopy with respect to hydrogen-1 nuclei within the molecules of a substance, in order to determine the structure of its molecules [1] . In samples where natural hydrogen (H) is used, practically all of the hydrogen consists of the isotope 1H (hydrogen-1; i.e. having a proton for a nucleus). Simple NMR spectra are recorded in solution, and solvent protons must not be allowed to interfere. Deuterated (deuterium = 2H, often symbolized as D) solvents especially for use in NMR are preferred, e.g. deuterated chloroform, CDCl3, and deuterated dimethyl sulfoxide, (CD3)2SO (DMSO). However, a solvent without hydrogen, such as carbon tetrachloride, CCl4 or carbon disulphide, CS2, may also be used. Historically, deuterated solvents were supplied with a small amount (typically 0.1 %) of tetramethylsilane (TMS) as an internal standard for calibrating the chemical shifts of each analyte proton. TMS is a tetrahedral molecule, with all protons being chemically equivalent, giving one single signal, used to define 0 ppm. It is volatile, making sample recovery easy as well. Modern spectrometers are able to reference spectra based on the residual proton in the solvent (e.g. the CHCl3, 0.01 % in 99.99 % CDCl3). Deuterated solvents are now commonly supplied without TMS. Deuterated solvents permit the use of deuterium frequency-field lock (also known as deuterium lock or field lock) to offset the effect of the natural drift of the NMR's magnetic field B0. In order to provide deuterium lock, the NMR constantly monitors the deuterium signal resonance frequency from the solvent and makes changes to the B0 to keep the resonance frequency constant.[2] Additionally, the deuterium signal may be used to accurately define 0 ppm as the resonant frequency of the lock solvent and the difference between the lock solvent and 0 ppm (TMS) are well known. Proton NMR spectra of most organic compounds are characterized by chemical shifts in the range +12 to -4 ppm and by spin-spin coupling between protons. The integration curve for each proton reflects the abundance of the individual protons. Simple molecules have simple spectra. The spectrum of ethyl chloride consists of a triplet at 1.5 ppm and a quartet at 3.5 ppm in a 3:2 ratio. The spectrum of benzene consists of a single peak at 7.2 ppm due to the diamagnetic ring current. Together with Carbon-13 NMR, proton NMR is a powerful tool for molecular structure characterization. Contents 1 Chemical shifts 2 Spin-spin couplings 3 Carbon satellites and spinning sidebands 4 See also 5 References 6 External links [edit] Chemical shifts Chemical shift values are not precise, but typical - they are to be therefore regarded mainly as orientational. Deviations are in ±0.2 ppm range, sometimes more. The exact value of chemical shift depends on molecular structure and the solvent in which the spectrum is being recorded. Hydrogen nuclei are sensitive to the hybridisation of the atom to which the proton is attached and to electronic effects. Nuclei tend to be deshielded by groups which withdraw electron density. Deshielded nuclei resonate at higher δ values, whereas shielded nuclei resonate at lower δ values. Examples of electron withdrawing substituents are -OH, -OCOR, -OR, -NO2 and halogens. These cause a downfield shift of approximately 2-4ppm at Cα and of less than 1-2 ppm at Cβ. Carbonyl groups, olefinic fragments and aromatic rings contribute sp2 hybridised carbon atoms to an aliphatic chain. This causes a downfield shift of 1-2 ppm at Cα. Note that labile protons (-OH, -NH2, SH) have no characteristic chemical shift. However such resonances can be identified by the disappearance of a peak when reacted with D2O, as deuterium will replace a proton. This method is called a D2O shake. Acidic protons may also be suppressed when a solvent containing acidic deuterium ions (e.g. methanol-d4) is used. Functional group CH3 CH2 CH CH2R 0.8 1.3 1.6 C=C 1.6 2.0 2.6 C≡C 1.7 2.2 2.8 C6H5 2.3 2.6 2.9 F 4.3 4.4 4.8 Cl 3.0 3.4 4.0 Br 2.7 3.4 4.1 I 2.2 3.2 4.2 OH 3.3 3.5 3.8 OR 3.3 3.4 3.7 OC6H5 3.8 4.0 4.3 OCOR 3.6 4.1 5.0 OCOC6H5 3.9 4.2 5.1 OCOCF3 4.0 4.4 / CHO 2.2 2.4 2.5 COR 2.1 2.2 2.6 COOH 2.1 2.3 2.6 COOR 2.0 2.3 2.5 CONR2 2.0 2.1 2.4 CN 2.1 2.5 3.0 NH2 2.5 2.7 3.0 NR2 2.2 2.4 2.8 NRC6H5 2.6 3.0 3.6 NR3+ 3.0 3.1 3.6 NHCOR 2.9 3.3 3.7 NO2 4.1 4.2 4.4 SR 2.1 2.5 3.1 SOR 2.6 3.1 / =O (aliphatic aldehyde) / / 9.5 =O (aromatic aldehyde) / / 10 M-H (metal hydride) / / [edit] Spin-spin couplings The chemical shift is not the only indicator used to assign a molecule. Because nuclei themselves are little magnets they influence each other, changing the energy and hence frequency of nearby nuclei as they resonate—this is known as spin-spin coupling. The most important type in basic NMR is scalar coupling. This interaction between two nuclei occurs through chemical bonds, and can typically be seen up to three bonds away. The effect of scalar coupling can be understood by examination of a proton which has a signal at 1ppm. This proton is in a hypothetical molecule where three bonds away exists another proton (in a CH-CH group for instance), the neighbouring group (a magnetic field) causes the signal at 1 ppm to split into two, with one peak being a few hertz higher than 1 ppm and the other peak being the same number of hertz lower than 1 ppm. These peaks each have half the area of the former singlet peak. The magnitude of this splitting (difference in frequency between peaks) is known as the coupling constant. A typical coupling constant value would be 7 Hz. The coupling constant is independent of magnetic field strength because it is caused by the magnetic field of another nucleus, not the spectrometer magnet. Therefore it is quoted in hertz (frequency) and not ppm (chemical shift). In another molecule a proton resonates at 2.5 ppm and that proton would also be split into two by the proton at 1 ppm. Because the magnitude of interaction is the same the splitting would have the same coupling constant 7 Hz apart. The spectrum would have two signals, each being a doublet. Each doublet will have the same area because both doublets are produced by one proton each. The two doublets at 1 ppm and 2.5 ppm from the fictional molecule CH-CH are now changed into CH2-CH: The total area of the 1 ppm CH2 peak will be twice that of the 2.5 ppm CH peak. The CH2 peak will be split into a doublet by the CH peak—with one peak at 1 ppm + 3.5 Hz and one at 1 ppm - 3.5 Hz (total splitting or coupling constant is 7 Hz). In consequence the CH peak at 2.5 ppm will be split twice by each proton from the CH2. The first proton will split the peak into two equal intensities and will go from one peak at 2.5 ppm to two peaks, one at 2.5 ppm + 3.5 Hz and the other at 2.5 ppm - 3.5 Hz—each having equal intensities. However these will be split again by the second proton. The frequencies will change accordingly: The 2.5 ppm + 3.5 Hz signal will be split into 2.5 ppm + 7 Hz and 2.5 ppm The 2.5 ppm - 3.5 Hz signal will be split into 2.5 ppm and 2.5 ppm - 7 Hz The net result is not a signal consisting of 4 peaks but three: one signal at 7 Hz above 2.5 ppm, two signals occur at 2.5 ppm, and a final one at 7 Hz below 2.5 ppm. The ratio of height between them is 1:2:1. This is known as a triplet and is an indicator that the proton is three-bonds from a CH2 group. This can be extended to any CHn group. When the CH2-CH group is changed to CH3-CH2, keeping the chemical shift and coupling constants identical, the following changes are observed: The relative areas between the CH3 and CH2 subunits will be 3:2. The CH3 is coupled to two protons into a 1:2:1 triplet around 1 ppm. The CH2 is coupled to three protons. Something split by three identical protons takes a shape known as a quartet, each peak having relative intensities of 1:3:3:1. A peak is split by n identical protons into components whose sizes are in the ratio of the nth row of Pascal's triangle: n 0 singlet 1 1 doublet 1 1 2 triplet 1 2 1 3 quartet 1 3 3 1 4 quintet 1 4 6 4 1 5 sextet 1 5 10 10 5 1 6 septet 1 6 15 20 15 6 1 7 octet 1 7 21 35 35 21 7 1 8 nonet 1 8 28 56 70 56 28 8 1 Because the nth row has n+1 components, this type of splitting is said to follow the "n+1 rule": a proton with n neighbors appears as a cluster of n+1 peaks. With 2-methylpropane, (CH3)3CH, as another example: the CH proton is attached to three identical methyl groups. The C-H signal in the spectrum would be split into ten peaks according to the (n + 1) rule of multiplicity. Below are NMR signals corresponding to several simple multiplets of this type. Note that the outer lines of the nonet (which are only 1/8 as high as those of the second peak) can barely be seen, giving a superficial resemblance to a septet. When a proton is coupled to two different protons, then the coupling constants are likely to be different, and instead of a triplet, a doublet of doublets will be seen. Similarly, if a proton is coupled to two other protons of one type, and a third of another type with a different coupling constant, then a triplet of doublets is seen. In the example below, the triplet coupling constant is larger than the doublet one. The analysis of such multiplets (which can get very much more complicated than the ones shown here) provides important clues to the structure of the molecule being studied. It should be emphasized that the simple rules for the spin-spin splitting of NMR signals described above only apply if the chemical shifts of the coupling partners are substantially larger than the coupling constant between them, otherwise there may be more peaks, and the intensities of the individual peaks will be distorted (second-order effects). [edit] Carbon satellites and spinning sidebands Occasionally, small peaks can be seen shouldering the main 1H NMR peaks. These peaks are not the result of proton-proton coupling, but result from the coupling of 1H atoms to an adjoining carbon 13 atom. These small peaks are known as carbon satellites as they are small and appear around the main 1H peak i.e. satellite (around) to them. Carbon satellites are small because Carbon 13 only makes up about 1% of the atomic carbon content of carbon, the rest of the carbon atoms are predominantly NMR inactive Carbon 12. Carbon satellites always appear as an evenly spaced pair around the main 1H peak. This is because they are the result of 1% of the 1H atoms coupling to an adjoined 13C atom to give a wide doublet (carbon 13 has a spin of a half). Note, if the main 1H-peak has proton-proton coupling, then each satellite will be a miniature version of the main peak and will also show this 1H-coupling, e.g. if the main 1H-peak is a doublet, then the carbon satellites will appear as miniature doublets, i.e. one doublet on either side of the main 1H-peak. Sometime other peaks can be seen around 1H peaks, these are known as spinning sidebands and are related to the rate of spin of an NMR tube. Carbon satellites and spinning sidebands should not be confused with impurity peaks [3] . [edit] See also Mass spectrometry Pople Notation – letter designations for spin-systems [edit] References ^ R. M. Silverstein, G. C. Bassler and T. C. Morrill, Spectrometric Identification of Organic Compounds, 5th Ed., Wiley, 1991. ^ US patent 4110681, Donald C. Hofer; Vincent N. Kahwaty; Carl R. Kahwaty, "NMR field frequency lock system", issued 1978-08-29  ^ Gottlieb HE, Kotlyar V, Nudelman A (October 1997). "NMR Chemical Shifts of Common Laboratory Solvents as Trace Impurities". J. Org. Chem. 62 (21): 7512–7515. doi:10.1021/jo971176v. PMID 11671879.  [edit] External links 1H-NMR Interpretation Tutorial Spectral Database for Organic Compounds Proton Chemical Shifts 1D Proton NMR 1D NMR experiment v â€¢ d â€¢ e NMR spectroscopy by isotope Isotope 1H - 2H - 13C - 15N - 17O - 19F - 29Si - 31P - 77Se - 195Pt References [1] || Proton–proton chain reaction From Wikipedia, the free encyclopedia Jump to: navigation, search This article needs additional citations for verification. Please help improve this article by adding reliable references. Unsourced material may be challenged and removed. (September 2009) The proton–proton chain reaction is one of several fusion reactions by which stars convert hydrogen to helium, the primary alternative being the CNO cycle. The proton–proton chain dominates in stars the size of the Sun or smaller. In general, proton–proton fusion can occur only if the temperature (i.e. kinetic energy) of the protons is high enough to overcome their mutual electrostatic or Coulomb repulsion. The theory that proton–proton reactions were the basic principle by which the Sun and other stars burn was advocated by Arthur Stanley Eddington in the 1920s. At the time, the temperature of the Sun was considered too low to overcome the Coulomb barrier. After the development of quantum mechanics, it was discovered that tunneling of the wavefunctions of the protons through the repulsive barrier allows for fusion at a lower temperature than the classical prediction. Even so, it was unclear how proton-proton fusion might proceed, because the most obvious product, helium-2, is unstable and immediately dissociates back into a pair of protons. In 1939, Hans Bethe proposed that one of the protons could beta decay into a neutron via the weak interaction during the brief moment of fusion, making deuterium the initial product in the chain.[1] This idea was part of the body of work in stellar nucleosynthesis for which Bethe won the Nobel Prize. In the Sun, deuterium-producing events are rare enough that a complete conversion of its hydrogen would take more than 1010 years at the prevailing conditions of its core.[2] The fact that the Sun is still shining is due to the slow nature of this reaction; if it went faster, the Sun would have exhausted its hydrogen long ago. Contents 1 The pp chain reaction 1.1 The pp I branch 1.2 The pp II branch 1.3 The pp III branch 1.4 The pp IV or Hep 1.5 Energy release 2 The pep reaction 3 See also 4 References [edit] The pp chain reaction The Proton–proton chain dominates in stars the size of the Sun or smaller. The first step involves the fusion of two hydrogen nuclei 1H (protons) into deuterium, releasing a positron and a neutrino as one proton changes into a neutron. 11H  +  11H  →  21D  +  e+  +  νe  +  0.42 MeV This first step is extremely slow, both because the protons have to tunnel through the Coulomb barrier and because it depends on weak interactions. The positron immediately annihilates with an electron, and their mass energy is carried off by two gamma ray photons. e−  +  e+  →  2 Î³  +  1.02 MeV After this, the deuterium produced in the first stage can fuse with another hydrogen to produce a light isotope of helium, 3He: 21D  +  11H  →  32He  +  γ  +  5.49 MeV From here there are three possible paths to generate helium isotope 4He. In pp I helium-4 comes from fusing two of the helium-3 nuclei produced; the pp II and pp III branches fuse 3He with a pre-existing 4He to make Beryllium. In the Sun, branch pp I takes place with a frequency of 86%, pp II with 14% and pp III with 0.11%. There is also an extremely rare pp IV branch. [edit] The pp I branch 32He  +  32He  →  42He  +  2 11H  +  12.86 MeV The complete pp I chain reaction releases a net energy of 26.7 MeV. The pp I branch is dominant at temperatures of 10 to 14 MK. Below 10 MK, the PP chain does not produce much 4He.[citation needed] [edit] The pp II branch Proton-Proton II chain reaction See also: lithium burning 32He  +  42He  →  74Be  +  γ 74Be  +  e−  →  73Li  +  νe  +  0.861 MeV  /  0.383 MeV 73Li  +  11H  →  2 42He The pp II branch is dominant at temperatures of 14 to 23 MK. 90% of the neutrinos produced in the reaction 7Be(e−,νe)7Li* carry an energy of 0.861 MeV, while the remaining 10% carry 0.383 MeV (depending on whether lithium-7 is in the ground state or an excited state, respectively). [edit] The pp III branch Proton-Proton III chain reaction 32He  +  42He  →  74Be  +  γ 74Be  +  11H  →  85B  +  γ 85B      →  84Be  +  e+  +  νe  +  γ 84Be      →  2 42He The pp III chain is dominant if the temperature exceeds 23 MK. The pp III chain is not a major source of energy in the Sun (only 0.11%), but was very important in the solar neutrino problem because it generates very high energy neutrinos (up to 14.06 MeV). [edit] The pp IV or Hep This reaction is predicted but has never been observed due to its great rarity (about 0.3 ppm in the Sun). In this reaction, Helium-3 reacts directly with a proton to give helium-4, with an even higher possible neutrino energy (up to 18.8 MeV). 32He  +  11H  →  42He  +  e+  +  νe  +  18.8 MeV [edit] Energy release Comparing the mass of the final helium-4 atom with the masses of the four protons reveals that 0.007 or 0.7% of the mass of the original protons has been lost. This mass has been converted into energy, in the form of gamma rays and neutrinos released during each of the individual reactions. The total energy yield of one whole chain is 26.73 MeV. Only energy released as gamma rays will interact with electrons and protons and heat the interior of the Sun. This heating supports the Sun and prevents it from collapsing under its own weight. Neutrinos do not interact significantly with matter and do not help support the Sun against gravitational collapse. The neutrinos in the ppI, ppII and ppIII chains carry away 2.0%, 4.0% and 28.3% of the energy in those reactions respectively.[3] [edit] The pep reaction Proton–proton and electron-capture chain reactions in a star. Deuterium can also be produced by the rare pep (proton–electron–proton) reaction (electron capture): 11H  +  e−  +  11H  →  21D  +  νe In the Sun, the frequency ratio of the pep reaction versus the pp reaction is 1:400. However, the neutrinos released by the pep reaction are far more energetic: while neutrinos produced in the first step of the pp reaction range in energy up to 0.42 MeV, the pep reaction produces sharp-energy-line neutrinos of 1.44 MeV. Both the pep and pp reactions can be seen as two different Feynman representations of the same basic interaction, where the electron passes to the right side of the reaction as an anti-electron. This is represented in the figure of proton–proton and electron-capture chain reactions in a star, available at the NDM'06 web site.[4] [edit] See also Triple-alpha process CNO cycle [edit] References Wikimedia Commons has media related to: Proton–proton chain reaction ^ Hans A. Bethe, Physical Review 55:103, 434 (1939); cited in Donald D. Clayton, Principles of Stellar Evolution and Nucleosynthesis, The University of Chicago Press, 1983, p. 366. ^ Kenneth S. Krane, Introductory Nuclear Physics , Wiley , 1987, p. 537. ^ Claus E. Rolfs and William S. Rodney, Cauldrons in the Cosmos, The University of Chicago Press, 1988, p. 354. ^ Int'l Conference on Neutrino and Dark Matter, Thursday 07 Sept 2006, http://indico.lal.in2p3.fr/getFile.py/access?contribId=s16t1&sessionId=s16&resId=1&materialId=0&confId=a05162 Session 14. v â€¢ d â€¢ e Nuclear processes Radioactive decay Alpha decay Â· Beta decay Â· Gamma radiation Â· Cluster decay Â· Double beta decay Â· Double electron capture Â· Internal conversion Â· Isomeric transition Â· Spontaneous fission Stellar nucleosynthesis pp-chain Â· CNO cycle Â· α process Â· Triple-α Â· Carbon burning Â· Ne burning Â· O burning Â· Si burning Â· R-process Â· S-process Â· P-process Â· Rp-process Other processes Emission Neutron emission Â· Positron emission Â· Proton emission Capture Electron capture Â· Neutron capture || Home Home About us Special offers News Find a dealer Contact us Gen-2 Savvy Satria Neo Ecologic Proton Cars (UK) Limited has made every effort to ensure the accuracy of the information contained in this site, but Proton believes in a policy of continuous improvement and therefore reserves the right to change, alter or modify, among other things, specifications, colours and prices of models without prior notice at any time. 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To contact Proton Customer Services directly: Proton Customer Services Proton Cars (UK) Limited 1-3 Crowley Way Avonmouth Bristol BS11 9YR 0844 8565 899 Special offers | Home | GEN-2 | Savvy | Satria Neo | Ecologic | About us | News | Find a dealer | Book a test drive | Request a brochure Finance | Media centre | Business | Legal | Site map | Contact us Proton Cars (UK) Limited | Registered Office: 1-3 Crowley Way, Avonmouth, Bristol, BS11 9YRRegistered in England No: 1857505 || Physics Forums > Physics > Quantum Physics > photon-proton/ proton-electron interactions PDA View Full Version : photon-proton/ proton-electron interactions no gunSep13-04, 07:53 AMhi i was just curious to find out how photons interact with protons (virtual and real) because as we know a photon excites an electron to a new state but what happens to a proton? also why dont electrons collide with the nucleus. i know they have acceleration which will keep them from "falling" into the nucleus however if 2 electrons come near each other and repel each other wont one spin closer to the nucleas (in which case its possible for it to spin straight in) while the other is repelled outwards? ive read two ways that electrons dont collide with the nucleas. one is that photons from the proton keep them excited enough not to come to close to the nucleus (which is why i asked the first question) and if this is the case what happens if a photon misses an electron (and i know electrons arent always particles (even though a recent experiment has proven that electrons exist in 2 states at once)). the other way that i know of that could be possible is that electrons is an actual electron cloud... a charge spread over an area. this is fine until i think of it being spread through the nucleus. if its a charge being spread then why isnt it attracted to the positive charge of the nucleus? and if it is an electron cloud then s p d f orbitals wont really exist would they? so any help would be nice thanks ps i know that schrodingers equation n cannot = 0 marlonSep13-04, 08:29 AMAn incident foton is absorbed by the target atom if it has just the exact amount of energy in order to trigger the excitation of the atom. This means that one of the electrons is pushed out of it's position leaving a hole. This is the excited state. A higher positioned (i mean higher qua energy) electron can fill this hole up by emitting some EM-radiation. This is how X-rays are generated. An electron will not interact directly with the atomic nucleus because of the surrounding electron cloud which yields a repulsive effect,hmm. The electron cloud is not attracted into the nucleus because the constituent electrons repell themselves. This leads to the equilibrium (attraction by the nucleus and repulsion by the other electrons) that leaves the cloud around the positive nucleus. This is just of the top of my head, i will post some more answers later on... till then regards marlon no gunSep13-04, 09:06 AMin answer This leads to the equilibrium (attraction by the nucleus and repulsion by the other electrons) that leaves the cloud around the positive nucleus there must be an electron right at the bottom which is nearest to the nucleus which will be pushed into the nucleus why doesnt this happen? and how do protons interact with photons? ZapperZSep13-04, 09:08 AMI will add just a little bit to what Marlon has said... hi was just curious to find out how photons interact with protons (virtual and real) because as we know a photon excites an electron to a new state but what happens to a proton? You need to make sure you understand this clearly. When a photon of the right energy hits an atom, it doesn't "excite" an electron, it excites the WHOLE ATOM. The energy states that the electron can occupy is a result of the combination of the nucleus and the electron, not just the electron alone. We know this to be true because a free electron has no such energy states. So the nucleus plays a significant role in forming those energy states for the electron to occupy. It is the atom that is excited upon photon absorption, not the electron. Zz. marlonSep13-04, 09:12 AMA cristal clear addendum from ZapperZ Indeed, the electron has the energylevels described by QM because it is in interaction with the atomic nucleus. regards marlon marlonSep13-04, 09:35 AMand how do protons interact with photons? This is an answer i gave in another thread. It describes how electrons interact with EM-waves (photons) due to the Lorentzforce. Every charged particle in an EM will feel this force, so the following explanation is just the same for protons. Just replace electrons by protons when reading The EM wave (photons) exerts a Lorentzforce onto the electron. This electron accelerates and the momentum goes from p to p''. Part of the momentum of the EM wave is being absorbed by the electron (Poynting vector). Because the electron is accelerated it will emit an EM wave with wavelength lambda'. So basically the incident photon has a wavelength that goes from lambda to lambda'. The momentum of the electron them changes into p'. So we have p --> p'' --> p' This EM way of thinking is a local fieldtheory because their is no activity of forces "on a distance". EM-forces are being carried over by fields that fill the entire "space" and they interact with charges positioned at a specific place. This is a big difference with the Newton-way of thinking. It is in complete accordance with the local fieldequations of Maxwell that electrons are pointcharges. This is a consequence of the fact that Maxwell equations need to be relativistically invariant. The only question remains as to why the entire EM-wave is absorbed as one single quantum. The answer to that question is ofcourse the wave-mechanics of Schrödinger...as we all know...(first quantization) So we have fields, and particles and it is the second quantization that gives us force carriers (viewed at as particles not as waves) and the fermionic matterfields (like the Diracfield being the general solution to the Dirac-equation)that yield the elementary massless particles of the Standard Model. regards marlon :cool: no gunSep13-04, 10:15 AMi still dont understand how the whole atom is excited when its just one electron rising a shell then dropping a shell once it releases the photon. the atom doesnt change just the electron. if a photon does not interact with a electron directly but does with a proton, will the fact that a proton is so large mean that it will just reflect the photon? also why doesnt the electron hit the nucleus? :P dat was da main question if the orbital of an electron is slightly changed due to repulsive forces from another electron wont that be enough for the electron to spiral into the nucleus (just like a satellite). in addition to the attractive force of the proton (which is quite large) so 3 question in summary :D 1) how do proton directly interact with photons (reflect or absorb?) 2) why dont the electrons collide with the nucleus in any case senario? 3) how is the whole atom effected when a single electron is excited and de-excited thanks :D ZapperZSep13-04, 10:36 AMi still dont understand how the whole atom is excited when its just one electron rising a shell then dropping a shell once it releases the photon. the atom doesnt change just the electron. Those "shell" that you talk about came about when you solve the schrodinger equation for a central, spherical potential. This is the NUCLEUS of the atom! Without the atom, there are no shells, no energy states, etc. There are no "absorption" of a photon by a free electron. There are scattering of a photon by free electrons. Thus, the electron in an atom are "incidentals". It is the whole atom that gained energy, not just the electron making the transition. When you exercise, you say that your whole body becomes warmer. Yet, technically, only certain parts of your body that are exerting most of the effect. Yet, you make no such distinction of what parts of your body are warmer and more "excited". You say your whole body systemically are warmer/hotter. Same thing with the atom.... Zz. no gunSep14-04, 04:33 AMso in other words the reason for an electron not hittin the nucleus is the fact that it is always excited because of constant photon interactions? and if we were able to stop these photons then what? marlonSep14-04, 06:50 AMso in other words the reason for an electron not hittin the nucleus is the fact that it is always excited because of constant photon interactions? and if we were able to stop these photons then what? I don't think so. The electron and the nucleus underdo many interactions. They are both charged so they we exert a Coulomb-interaction onto eachother. If you would look at the corresponding potential, you will see that this potential has a minimal value when plotted in function of the distance r between the two interaction particles. This minimum in potential energy denotes the binding-state because everything in nature will strive to be in the state of lowest potential energy. You could say that nature is as lazy as possible. Now, the distance r corresponding to this lowest potential energy is NOT 0 but has a certain finite radius (the Bohr-radius for example) , thus the electron has a minimal distance to the nucleus that is equal to this r-value. Keep also in mind that two things have to be taken into account. First you have the "attraction" between the electron-cloud and the nucleus, but the electrons themselves are repelled because of the many electrons that make up the cloud surrounding the nucleus. An atom does not need photons hitting it in order to be in this stabe equilibrium. Finally, stopping a photon (or changing it's path) would require extern magnetic fields in order to exert a Lorentz-force on it. ofcourse this also brings a change in the energy-levels of the atom (e.g. The Starck and Zeemann-effect). Let's not do that... :smile: regards marlon no gunSep15-04, 10:04 AMok thanks ... still doesnt seem to logical to me but i guess ill have to deal with it marlonSep15-04, 10:52 AMWhat is it, that maken this look unlogical to you... I will try to explain to you... regards marlon pmokoFeb28-08, 08:00 PMHello, I got the same question. But for simplicity, lets talk about hydrogen atom, with one proton and one electron (so there is not electron cloud, just 1 electron). Why electron doesn't collide with nucleus (proton), but keeps orbiting around it? (There is an acceleration, but lets assume that sooner or later it is gone be externaly disrupted in a way that electron is gone be pushed towards nucleus). marlonMar1-08, 07:37 AMHello, I got the same question. But for simplicity, lets talk about hydrogen atom, with one proton and one electron (so there is not electron cloud, just 1 electron). Why electron doesn't collide with nucleus (proton), but keeps orbiting around it? (There is an acceleration, but lets assume that sooner or later it is gone be externaly disrupted in a way that electron is gone be pushed towards nucleus). Check out the answer to your question : http://www.physicsforums.com/showpost.php?p=862093&postcount=2 marlon Codybear1977Jul20-09, 05:16 AMThe electron cloud is not attracted into the nucleus because the constituent electrons repell themselves. This leads to the equilibrium (attraction by the nucleus and repulsion by the other electrons) that leaves the cloud around the positive nucleus. How does this relationship work in say, Hydrogen, when there is only one electron? If the electron exists as a smeared cloud, precluding it from contacting the nucleus indefinately, then neutron stars could not exist, as this process of annihilation between the protons and electrons would be necessary for that. If the electrons do not exist as a solid tangible mass, succeptible to the laws of physics, how do they succumb to pressure and gravity, eventually "falling" into the nucleus in the collapse of a neutron star? Bear in mind I am not a physicist, so please forgive me if my questions seem unintelligent. vBulletin® v3.7.6, Copyright ©2000-2010, Jelsoft Enterprises Ltd. || Myspace Myspace Beta Entdecken▼ Loading... Registrieren • Einloggen▼ Das neue Myspace Kostenlose Musik • Kostenlose Games • Kostenloses Fernsehen Kostenlos registrieren Schon Mitglied? Logge dich ein. E-Mail Passwort Eingeloggt bleiben Passwort vergessen? Einloggen Suchen Start Leute Musik Video Games Events Mehr â–¼ Comedy Forum Gruppen Schulen/Unis Handy Designs Was gibt's Neues? 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