Show that if an electron were confined inside an atomic nucleus of diameter 2.0 x 10^-15 m, it would have?
350188
The proper way to do this would be to calculate the quantum mechanical energy levels for an electron in a spherical well with potential = 0 for r < 10^-15 m, and infinite potential for r >= 10^-15m (ignoring relativistsic effects for now).  Because, by definition, the potential energy of the electron is zero everywhere within this spherical "box", all the energy is kinetic energy.  You can then calculate what the classical velocity would have to be in order for the electron to have such a kinetic energy.  You should find that the answer is an appreciable fraction of c (if not in excess of c).  This tells you that you need to consider relativistic effects.\n\nA back-of the envelope calculation to show the same thing invokes the uncertainty relationship, delta-x * delta-p > h/4*pi, where delta-x is the uncertainty in position, delta-p is the uncertainty in momentum, and h is Planck's constant.\n\nIn this case, by definition, the uncertainty in position is +/- 10^-15*m.  Plugging this into the uncertainty relationship and solving for delta-p yields:\n\ndelta-p = 5.3 * 10^-15 gm*cm/s\n\nClassically, the momentum is equal to m * v, so dividing the momentum uncertainty by the rest mass of an electron (9.1 * 10^-31 kg) yields:\n\ndelta-v = 5.8 * 10^10 m/s,\n\nwhich exceeds c, telling us that we need to go back and take relativistic effects into account.
544865
depends on how much an electron has been drinking
304768
The assembled nucleus has less energy than the collective nucleons. If it did not\n\na) it would not hold together with the collection of + charged protons, and\n\nb) it would not require added energy in the form of colliding particles to make it come apart.
309704
Approximately 193,000 atoms thick
791870
The size of the atom decreases as you move right.  As you move across the period you fill up the s and p orbitals.  These are then pulled closer to the nucleus.
512153
The mean van der Waals diameter of water has been reported as identical with that of isoelectronic neon (2.82 Å) [112]. Molecular model values and intermediate peak radial distribution data indicates however that it is somewhat greater (~3.2Å). The molecule is clearly not spherical, however, with about a ±5% variation in van der Waals diameter dependent on the axis chosen; approximately tetrahedrally placed slight indentations being apparent opposite the (putative) electron pairs.
97906
No electron path. If this was true electrons would have eventually collapsed into the nucleus
742739
The binding energy of an individual electron in Cs metal is given by the molar work function divided by Avogadro's number:\n\nE_e = 184 kJ/mol / 6.022*10^23 mol^-1\nE_e = 3.055*10^-19 J\n\nThe energy of a photon with wavelength 0.628 nm is given by:\n\nE_p = h*c/lambda\n\nwhere h is Planck's constant\nc is the speed of light in a vacuum\nlambda is the wavelength\n\nso, in this case,\n\nE_p = (6.626*10^-34 (kg*m^2)/s * 2.998*10^8 m/s)/ 0.628*10^-9 m\n\nE_p = 3.163*10^-16 J\n\nE_p exceeds E_Cs by a large margin, so the incoming photon has more than enough energy to eject an electron from the material.  Any energy imparted to an electron that is in excess of the amount needed to eject the electron from the metal will show up as kinetic energy of the electron.\n\nAn electron in a Cs atom beneath the surface of the matrial will require more energy to be ejected than one closer to the surface because the subsurface electron needs additional energy to make its way to the surface.  A electron on a Cs atom at the very uppermost surface will require the least energy to be ejected, and that energy is equal to its binding energy, and it will therefore have the largest "excess" energy, largest kinetic energy, and highest speed.\n\nIn this case, the "excess" energy (Call it E#) is:\n\nE# = E_p - E_e = 3.163*10^-16 J - 3.055*10^-19 J = 3.160 * 10^-16 J\n\nThis energy shows up as kinetic energy:\n\nE# = 0.5 * m * v^2\n\nwhere m is the mass of an electron\nv is the speed of the electron\n\nIn this case:\n\n3.160 * 10^-16 J = 0.5 * 9.109*10^-31 kg * v^2\n\nv = 2.634*10^7 m/s  (I get a slightly different number than the one you were given, which I attribute to rounding differences)\n\nThis speed is 2.634*10^7/2.998*10^8 = 0.088 = 8.8% that of the speed of light in a vacuum.
495461
Atomic radius (or size) increases as you go down and to the left on the periodic table.  Therefore, out of these, the largest would be K.
45899
Start thinking of electrons in an atom as probability clouds...or regions of space around the atom at a given radius where the probability is high that the electron will exist.  Bohr theory ("orbits") is not complete, in fact it's not an accurate depiction of the electrons in an atom.\n\nThe reason the electron's don't radiate while "orbiting" is that they don't "orbit" anything!!  Instead, there are regions of space around the atom where the probability of an electron existing at any instant is non-zero.  Excited electrons may lose energy and drop to a lower energy level (to a smaller radius from the atom), in which case the electron will radiate energy in the form of a photon whose energy is equal to the difference in energy levels (excited state - lower state).  \n\nAs a concrete example, this radiation from dropping to lower energy states is the basis of lasing action and the principle on which lasers operate.\n\nAs I said, think in terms of the quantum mechanical description of the electron rather than the Bohr theory.  It's not over your head, believe me, just study quantum mechanics.  Start with the electron in a Hydrogen atom, it's the simplest case (a 1 electron atom!!)\n\nI hope this helps.
52468
it is the charge of an electron 1.602*10^-19 Coulombs, i think that is what you are referring to.
127803
Neils bohr suggested that the nucleus was positively charged and the electrons were negatively charged. The electrostatic force between them kept electrons in their orbits. The electrons moved in specific orbits only. They did not radiate energy continously but in the form of many quanta (packets) of energy. Electron emitted energy only when it jumped from a higher orbit to a lower orbit.\n\nMathematics of Bohr's atomic model :\nLet charge on proton be e and that on electron be -e. Let radius of the nth orbit be r. The electrostatic force between nucleus and the electron provides the electron a necessary centripedal force. Let the velocity of the electron be v and mass be m.\n\n m(v^2)/r = k(e^2)/(r^2)    where k = 1/(4*pi*epsilon)\n                         epsilon is permittivity of space \n\nThis is the first postulate of his theory.\n\nElectrons move only in those orbits whose angular momentum is an integral multiple of h/2pi where h is plank's constant\n\n  mvr = Iw = nh/(2*pi)\n\nI - moment of inertia of electron\nw - angular velocity of electron.\n\nThis is the second postulate.\n\nLet the energy of the initial orbit of electron be E2 and the energy of final orbit be E1. Then when it jumps from initial orbit to the final orbit, the energy liberated is in the form of radiations having frequency f and is numerically equal to the difference in energies of the two energies of electron in the two orbits.\n\n                dE = E2 - E1\n\nHence,          hf = E2 - E1\n where h is plank's constant.\n\nThis is the third postulate.\n\nI hope this solves your query. Post your queries if you feel unsatisfied.
658056
The Schwarzschild radius(for those who want to be a pain in the butt, yes this is a simplified equation for a nonrotating mass but for our purposes should be a good approximation) is 2GM/c^2 where G is the gravitational constant 6.67e-8 dyne cm^2 g^-2, M is the mass of the object, and c is the speed of light(use cm/s^2 as units).  Basically when all the mass is contained within a radius smaller than this value, the gravitational metrics go a little crazy and this is defined to be a black hole.\n\nI leave it to you to do the math, see whether the accepted radius of a nucleus is within this radius or not and there is you answer.  There are other forces at play that cause the strong attractive forces within a nucleus, particularly strong nuclear force.  If this was gravity(which from above you should see the radius is a lot smaller than the radius of the nucleus in order to cause such effects), then the electrons would be sucked into nucleus as well, since you would have both an attractive coloumb force and an attractive strong gravitational force.\n\nThat being said, you would not see radiative activity if the nucleus was a black hole.
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well, the place to start is the periodic table.  uranium's configuration number is [Rn] 5f3 6d1 7s2, which means the outermost electrons are in the n = 7 shell.\nto calculate the radius of 7s electrons requires some seriously ugly quantum mechanics, so i'll just present you with some results:\nhttp://web.mit.edu/3.091/www/pt/pert1.html\nwhich states that the atomic radius is 1.38 angstroms.  other places on the web list 1.45 angstroms and 1.7 angstroms, so about 1.5 seems like a pretty good average.  (an angstrom is 10^-10 m, or 0.1 nanometer)\n\nas for the nuclear radius, this has to be measured from proton-scattering experiments.  typically they are around 1 femtometer, which is 10^-15m or 0.00001 of an angstrom.  (see the wikipedia page for more detail)\n\ncheers!
1138753
This is a natural question if you think in terms of classical physics.  In fact, it's an even bigger problem than at first glance because, in classical physics, a charged object which accelerates will radiate energy.  So, in classical physics, all electrons would be expected to spiral in to the nucleus.  This puzzle cannot be addressed without quantum mechanics.\n\nHere's a very simple explanation.  It's not mathematically exact, but it carries the important concepts, and does give an answer fairly close to the mathematically correct one.\n\nFor simplicity, we'll consider hydrogen.  A single proton, no neutrons, and one electron.  There are no competing electrons to push each other apart, it really looks as if the best situation is for the electron to be squeezed right into the proton.\n\nRecall that in quantum mechanics, particles behave like waves.  Let's look at the electron that way.  What's the wavelength of this electron wave?  Well, we can't really say exactly, but a good approximation would be the wavelength is about the same as the width of the space in which you constrain the electron.  Put the electron in a large space, and it has a large wavelength.  Put it in a tiny space, and its wavelength is much smaller.  Ah, but remember, the shorter the wavelength, the higher the frequency, and the higher the frequency, the higher the energy.\n\nNow, you have two competing effects.  If the electron goes in very close to the proton, and stays there, then it will be lowering its electrostatic energy.  But to do that, it has to confine itself to an extremely small volume, which will give it a very high "wavelength energy".  It seems natural that there's a middle point somewhere such that if the electron got closer to the proton, its wavelength energy would increase faster than its electrostatic energy would decrease.  This sets a minimum distance for the electron from the proton.  It actually needs more energy to get closer to the proton after that.  So, the place the electron winds up is actually its lowest energy state, it's stable there, and doesn't require any injection of energy to maintain this position.\n\nAs I said, this is very non-mathematical, but it gives you an idea of what's happening in the quantum mechanics which explain the effect.
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The answer comes from quantum mechanics and involves some graduate level math. \n\nThat being said the general rule is that a Principal quantum number can have up to 2n^2 electrons.
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It would take about 57000 Earths to span the classical diameter of an electron.\n\nThe universe has been in existence for 1.37E+10 years.  It would therefore be somewhere around that many lightyears in radius.  That's 1.296E+23 kilometers.\n\n(In fact, the universe is probably larger than this because of the inflationary expansion during which space appeared as a consequence of the relaxation of the primordial energy that created the universe to begin with.)\n\nEarth's radius is 6378 kilometers, which means that it would take 2.032E+19 Earths to span across the universe.\n\nOn a scale at which the universe is one meter wide, Earth's radius would be 4.921E-20 m.  The classical radius of an electron is 2.82E-15, so the electron (in the real world) would be (classically speaking) 57000 times wider than Earth (in the model).\n\nHowever, the Planck length is 1.616E-35 meters, and so even the scaled down Earth is still 3.045E+15 Planck lengths in radius.  If the Planck length were scaled UP by the same amount, it would still be only about 21 Angstroms long: about the wavelength of x-rays.