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Problem 112 from chapter 8.
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In this problem, they tell us that the first ionization energy of sodium is 496 kilojoules of mole.
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Use kulam's law to estimate the average distance between the sodium nucleus and the 3s electron.
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How does this distance compare to the atomic radius of sodium? so first, for kulom's law, we need a force or energy, which we can obtain from the, ionization energy.
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So this energy here will be equal to the negative ionization energy, which we're told is negative 5, 490 or 496 kilojoules per mole.
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But we can also manipulate these units here.
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So we're interested in the electrons, so we don't, per electrons, so we don't want to know the mole, we want to know the particle.
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So we divide by avogadra's number.
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And then we also convert from kilojoules to joules and we get a value of negative 8 .24 times 10 to the negative 19 joules for atom for this ionization energy which will be equal to the force from cool so we know kulums law, we have this force or energy equal to this constant times the two charges multiplied together times divided by the radius.
01:54
So we can rearrange this to solve for the radius since that's what we're looking for.
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And so when we do that, we get k -e -q -1, q2 divided by this energy, which we calculated above.
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And so that's equal to our constant, which is 8 .998 times 10 to the 9th, q1, which will be the charge of the nucleus.
02:25
So we multiplied the charge of a proton times 11, because that's how many protons are in this sodium nucleus...