What is the calculation for the energy emitted or absorbed when an electron changes energy levels?
374796
In a hydrogen atom, the energy of each level is given by:\n\nE=(-Z^2)(E0)/n^2\n\nwhere Z=the charge on the nucleus, E0=the energy of the ground state (the lowest level), n=an integer corresponding to the state (it equals 1 for the ground state.)\n\nTo find the energy emitted or absorbed, you simply subtract the energies between the two levels.\n\nI don't think you're likely to encounter any atoms more complicated than hydrogen or positive helium ions (which act like hydrogen). In any case, if you need the more general formula, don't be afraid to e-mail me. The above will work, by the way, for any atom that can be modelled with the Bohr model - it is actually just what the Bohr model gives for the energy of each level.
52468
it is the charge of an electron 1.602*10^-19 Coulombs, i think that is what you are referring to.
175956
C, remember that the electron loses and gets energy several times through the chain.
627895
When an electron goes from lower energy orbital to higher energy orbital it absorbs energy and when an electron drops from a higher energy orbital to a lower energy orbitaltehn it emits energy.\n\n(a) Emission (read the reason above)\n(b) Absorption (Read reason above)\n(c) Emission (read reason above)\n(d) Absorption (read reason above)
817827
energy levels/orbitals
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The energy of the incoming photons is given by Planck's relationship:  energy = Planck's constant * frequency.\n\nYou are given the wavelength, not the frequency, so you need to know that for any wave, the wavespeed = frequency * wavelength, so frequency = wavespeed/wavelength.  For photons, wavespeed = c.\n\nh = 6.626068 × 10^-34 34 m^2 kg / s (Planck's constant)\nc = 2.998 × 10^8 m / s (speed of light)\nwavelength = 48.2 x 10^-9 m (given in problem)\n\nPutting this together, the energy of each incoming photon is:\n\nE_photon = h*c/(48.2 x 10^-9 m) = 4.121 x 10^-18 J\n\nThe outgoing electron has kinetic energy = 1/2 * m * v^2.  You are given both m and v:\n\nE_electron = 1/2 * 9.109 x 10^-31 kg * (2.371 x 10^6 m/s)^2 \nE_electron = 2.56 x 10^-18 J\n\nThe ionization energy is the difference in energy between the incoming photon and the kinetic energy of the outgoing electron.  Because the question asks for the energy in joules per mole, we must multiply the energy difference for a single photon/electron pair by Avagadro's number:\n\nE_ionization = (E_photon - E_electron)* N_A\n\nE_ionization = (4.121 x 10^-18 J -2.56 x 10^-18 J)*6.022 x 10^23 mol^-1\n\nE_ionization = 9.4 x 10^5 J/mol = 940 kJ/mol
1135911
The photoelectric effect is the emission of electrons from matter upon the absorption of electromagnetic radiation, such as ultraviolet radiation or x-rays. An older term for the photoelectric effect was the Hertz effect, though this phrase has fallen out of current use.
576205
Electrons are in orbitals of a set quantum energy levels. There are normally empty orbitals above the filled orbitals. When an electron is excited to a higher energy level it does not want to stay there it want back to the lowest energy level, or at least a lower energy level. When it drops into the various possible energy levels it releases light at various frequencies equal to the energy of the electrons drop plus or minus any coupling. Hydrogen spectrum is the various frequencies of light that can be emitted by this process from hydrogen. It is solvable using Schrodinger's Equation.
1155911
Photoelectron energy (E) is given by\n\nE = W - hc/L\n\nWhere W is the work function of the metal and L is the excitation wavelength.
794988
Energy radiated per second is given by\nE = e*sigma*A*T^4   where \ne = emissivity; \nsigma = Stefan's constant = 5.67 * 10^-8 W/m^2/K^4\nA = surface area\nT = temperature in kelvin.\nSubstituting the various values,\nE = 249 * 10^-8 Watts.
1170187
The absorbed photon may be re-emitted as either reflected or refracted light. If it is absorbed, its energy is converted into heat. In some materials, a high-energy photon will stimulate the emission of a lower-energy photon. This is called fluorescence.
1110955
A pattern of dark lines or bands superimposed on a continuous spectrum. When a continuous spectrum of radiation (a broad range of wavelengths) passes through a material medium (for example, a cool, low-pressure gas), selective absorption occurs at certain particular wavelengths. This gives rise to a series of dips in intensity (absorption lines) which, in the visible region of the spectrum, appear as dark lines against the bright background of the ‘rainbow’ band of colors that comprises the continuous spectrum. Photons of electromagnetic radiation may be absorbed through radiative excitation, a process that occurs when an electron in one of the lower energy levels of an atom or ion absorbs a photon with energy precisely equal to the difference in energy between that level and one of the higher permitted levels and, as a result, jumps (makes an ‘upward transition’) from the lower to the higher level. Because the energy of a photon is inversely proportional to wavelength, the larger the energy gap, the shorter the wavelength of the radiation that is absorbed when an electron makes a particular transition. An electron does not normally remain in an excited level for long (typically 10−8 s). When it drops down again, it emits a photon or photons that may or may not have the same wavelength as the one that was originally absorbed (depending on whether it drops directly to its original level or descends in a series of smaller steps). If it drops in a series of steps, each of which corresponds to the emission of a photon of lower energy (and longer wavelength) than the one thatwasoriginally absorbed, the total number of photons of the original energy will be reduced and the spectrum will be depleted at the input wavelength. Furthermore, although all the input photons were traveling along essentially the same direction (from the source to the observer), the re-emitted photons travel away in random directions. Consequently, far fewer photons at the absorption wavelength reach the observer than photons of other wavelengths. The resulting absorption line is darker than the adjacent part of the continuous spectrum but, because some photons of that wavelength do reach the observer, not totally black.\nBecause each permitted transition corresponds to absorbing light of one particular wavelength, atoms or ions of a particular element produce absorption lines at a number of different wavelengths, each chemical element having its own distinctive ‘fingerprint’ pattern of absorption lines. The shortest-wavelength lines correspond to the largest energy gaps (i.e. to transitions from the lowest, or ‘ground’, level of the atom or ion). If an electron absorbs a photon with energy in excess of the ionization energy (or ‘ionization potential’), it will be removed from the atom. The absorption spectrum of a particular species of atom consists of several series of lines, corresponding to the various permitted transitions, the short-wavelength limit of the series corresponding to the photon energies above which ionization takes place. Because ionizing photons can, in principle, have any value of energy above the ionization energy, absorption can take place over a continuous range of wavelengths, shorter than the series limit. Absorption of this kind is called an absorption continuum. The prominence (‘strength’) of any particular line depends on the number of atoms of the appropriate chemical element that have electrons residing in the energy level from which the relevant upward transition takes place (the degree of excitation). That, in turn, depends on the abundance of the particular chemical element (the relative proportion of that element in the absorbing substance) and on a number of other factors, in particular the temperature (the higher the temperature, the greater the proportion of electrons in excited states).\nIn addition to producing absorption through electronic transitions (like atoms and ions), molecules may also a
1115839
No, the electron's charge is not its energy content, the charge is the charge which is a fundamental quantity (and the same for all electrons.) The volt is not a unit of force, it is a unit of electrical potential, with units of J/C (joules/coulomb). A volt represents a potential difference that will impart one joule of energy to one coulomb of charge passing through it. An electron volt is a unit of energy, defined as the amount of energy imparted to one electron passing through a potential difference of one volt. Since the electron's charge is −1.60217653*10^-19 C, the energy so imparted is equal to 1.60217653*10^-19 J.
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An electron passing through a potential difference of 150 kV picks up 150keV of energy.  So:\n150keV=h*c/lambda\nlambda =h*c/150keV.\nh=6.626*10^-34 J*sec\nc=3*10^8m/sec\n1eV=1.6*10^-19 J\n\nSo, lambda=((6.626*10^-34)(3*10^8))/((150000)(1.6*10^-19))\n\nlambda=8.28*10^-12 m, or 8.28 nm
919041
We're basically looking for a ratio of ionization energy E(n), where n =4 and n = 1 (ground state). Electron energy levels of a hydrogen atom is given by:\n\nE(n) = 13.6/n^2 eV\n\nTherefore,\n\nE(4)/E(1) =  1/16\n\nIt takes only 1/16 of 13.6 eV , or 0.85 eV, to ionize the ground state electron from a hydrogen atom on this very hot planet.
875956
The energy of a photon is proportional to its frequency, the constant of proportionality being called Planck's constant who discovered this relation.\n\nE = h * Frequency\n\nIf you know the frequency (or wavelength) of the light you can calculate the corresponding energy.\n\nWhen an electron jumps from an excited state back to its ground state it will emit energy.  If the frequency is in the right range we will see that as light.
1118486
Suppose a charged particle moves at a more or less constant velocity v1 followed by a finite time interval t of acceleration resulting in a new velocity v2. The radiation emitted always spans a large range of frequencies which (as a rough approximation) goes from frequencies of about 0 to 1/t, dropping to zero at frequencies much higher than 1/t.\n\nFor the acceleration events in your examples: For an electron passing through a sharp local magetic field, t ~ a/v where a is the size of the region with the magnetic field and v the electron velocity. For a collision with a nucleus or other charged particle t ~ b/v where b is the impact parameter (the closest distance between the straight line trajectory of the electron and the particle it's colliding with).\n\nFor purposes of order of magnitude calculations, the total energy per frequency interval radiated by a particle with charge e (in gaussian cgs units) is roughly:\n\nd Energy/d omega  ~ e^2 * |v2-v1|^2/c^3\n\nfor frequencies in the region 0 < omega < 1/t.\n\nThere is never one single frequency emitted except when there's a sustained pure sinusoidal varying current as in a radio antenna. Of course atomic and nuclear transitions produce photons with precise frequencies (wavelengths) but those are not really the result of accelerating charges in the usual sense.