The amount of energy necessary to remove an electron from an atom is a quantity called the ionization energy, Ej. This energy can be measured by a technique called photoelectron spectroscopy, in which light of wavelength Iv is directed at an atom, causing an electron to be ejected. The kinetic energy of the ejected electron (Ek) is measured by determining its velocity, v (Ek= mv2/2), and Ei is calculated using the conservation of energy principle. That is, the energy of the incident light is equal to Ej plus Ek. What is the ionization energy of selenium atoms (in kilojoules per mole) if light with wavelength = 48.2 nm produces electrons with a velocity of 2.37l x 10e6 m/s? The mass, m, of an electron is 9.109 x 10e-31 kg. (Remember 1 Joule= 1 kgm2/s2)
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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
352906
The light you see comes from photons emitted when an electron drops from a higher-energy orbit to a lower-energy one.  Quantum mechanics requires that the length of the orbit be an integral multiple of the wavelength of the electron's wavefunction.  That means these orbit energy levels are discrete, and so the energy difference between the levels has a very specific value, and the photon emitted must have exactly that energy and therefore that frequency.  Different elements will have different possible orbital energy levels, and therefore different emission spectra.
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The atoms' electrons absorb the energy (usually from the heat [flame test]). When they do this the extra energy causes the electrons to jump to the next energy level. They (the electrons) then loose the energy and fall back. In doing so give of a (depending on your book, or outlook) wave length of a certain frequency (light color). Or a photon of a certain color.\n\nGet an "A".
501054
The smaller the atom, the harder it is to remove an electron from it. This is the measure of an atom's ionisation energy.\n\nThe smaller an atom (with the exception of the noble gases),  the more likely it is to accept an additional electron. This is a measure of its electron affinity.\n\nBoth of these concepts factor in when determining whether bonding between two atoms is covalent or ionic.
142376
It depends on the Voltage.   Imagine an electron (which has known charge q = 1.6x10^-19 and mass m = 9.11x10^-31) that is inside a wire which is a perfect conductor.   Since the electron will move in a perfect conductor, all the potential energy given by the power source (Voltage * electron charge) will transform into kinetic energy (K = (m*v^2)/2), so:\n\n(m*v^2)/2 = V*q     ;     where v is the speed of the electron and V is the voltage\n\nWith this we get:   v = (2*V*q/m)^(1/2)\n\nAll the electrons would move at this same speed, and so the "speed of electricity" is given by the above formula.   For a voltage of 110V, the kind we have at home, the speed would be:\n\nv = 6'216,016.3 m/s\n\nCompared to the speed of light:\n\nc = 300'000,000 m/s\n\nThis analysis is for didactic purpouses, since for such high speeds we would need to consider relativistic effects.   Also, remember that we don't have perfect conductors, so the actual "speed of electricity" at 110V would be a lot less than the value shown above.\n\nI hope this answer was good enough.   :)
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A. You first begin with the relevant chemical equation ie:\n\nAl^3+ +3e^- --> Al\n\nnow work out how many moles of Al is in 1000 kg. From the above equation, you can see that each mole of Al requires 3 moles of e-s. Use Faraday's constant to find how many coulomb's of electricity is required. \n\nB. Energy = QV\n\nwhere:\nQ = charge in Coulombs (from part A above)\nV = voltage (ie 5.0 in this question).\n\nC. To calculate the voltage of an electrochemical cell, use the formula:\n\nEo cell = Eo cathode - Eo anode. \n\nIf you do not have a table of standard reduction potentials, you can readily find one by googling it.\n\nI hope that the above helps. If you need more help, post again with details of which part you don't understand.
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The change of energy  between energy levels.\n\nEnergy of electron in it's higher state  - energy of electron in it's lower state = h*f\n\nI forgot how to calculate the electron's energy at a specific energy state, but it should be easy to find.
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Indeed; this has to do with atomic orbital structure. Atoms try to fill their outermost electron shell with electrons. All atoms will release energy when they accept an electron (this energy is called its electron affinity) and absorb energy when they free an electron (called the ionization potential). Atoms that are a few electrons short of filling their last shell (such as O, F, Cl and most of Group IIIA-VIIA on the periodic table) "will do anything" to fill this last shell, and will release a great amount of energy in the diatomic form (O2). Thus two single O atoms are not stable very long, they would immediately react to form O2 which is far more stable. O is so unstable it would even react with O2 to form O3 if it can't find another O atom.\n\nThe fact that some atoms "exist only" in diatomic form in nature does not mean it is impossible to single out an O atom. If you bombard O2 atoms with light that has a higher energy than the bond, you will break it. And, for a brief period of time, will have to single O atoms which will very quickly react with one another, or with another molecule to achieve stability.\n\nHope that helps; good luck with your studies.
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E=hf \nwhere,\nh=6.63e-34 (planck's constant)\nE=(6.63e-34)(2.56e3)\n =1.7e-30\nCalculate for photon\n(1000j/sec)x(1 photon/1.70x-30)\n=5.88e32 photons/sec
279213
For this you should see the electronic configuration of those elements.\nThe metals with atomic number 21-30 are transition metals; they have outer 4s shell complete and the inner 3d orbital is 'in the process' of being filled up. They all have 2 outermost electrons in the 4s2 shell. \nIonization energy is the energy required in removing an outer most electron from a neutral atom. The transition metals and calcium have their outermost 2 electrons in 4s which means that their electrons experience the same force of attraction from the nucleus (being at the same distance from the nucleus) ; hence almost the same amount of energy will be required to overcome that attractive force to remove the electrons.
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E = mc^2 = 1kg * (3*10^8)^2 = 9^10^16 Joules
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E = energy\nm = mass\nc^2 = speed of light squared\nMass and energy are equivalent.\nThis is a part of theory of relativity of Albert Einstein.\n\nDuring nuclear fission and fusion about 1% mass disappears and an equivalent amount of energy comes free.\n\nDuring annihilisation all mass 'changes' into energy.\nDuring creation of an electron and positron the inverse happens.
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Electrons absorb energy to move from a lower level to a higher one.  They release energy going from higher to lower.
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The energy contained within a single photon of light can be found using Planck's equations,\nE = hv\nWhere E is the energy, h is Planck's constant, and v is the frequency of the light.\n\nYou can see that the higher the frequency of light, the higher the energy per photon.\nExamining the spectrum of light one can see that green light (roughly a wavelength of 550 nm) has a higher frequency than orange light (roughly a wavelength of 650 nm).\nTherefore, green light photons contain more energy than orange light photons.\n\nPlanck's constant = h = 6.626 E-34 J s
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No electrons will be emitted.\n\nIn order for the electron to jump to a higher level, or for that matter, be removed all together, they must absorb a specific minimum amount of energy...anything less than this and nothing will happen.\nIf it takes 4 eV to remove the electron and it only absorbs 3 eV, the electron will not be removed.
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There is potential energy and kinetic energy.  Kinetic energy is the energy that something has from motion.  It's equation is E=1/2*m*v(squared) where m=mass and v=velocity.  For potential energy there are many forms.  Gravity E=mass*height*gravity, Electric and magnetic fields are potential energy.  And nuclear energy E=m*c(squared) which is the rest potential energy that anything with mass will have.  There is also energy from light E=h*nu, nu=frequency h=planck's constant.  There is also chemical potential energy which is the energy contained within molecules.
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Kinetic energy is .5 * mass * velocity ^ 2\nSo the change in kinetic energy, or delta K.E, would be:\n\n.5 * mass * initial-velocity ^ 2 - .5 * mass * final-velocity ^ 2\n\nSince Kinetic Energy is lost, the initial velocity goes first.\n\nGood luck on your test!