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Bohr radius of a helium ion
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[QUOTE="leroyjenkens, post: 4528450, member: 162373"] Helium has a greater positive charge, so the nucleus going to pull on the lone electron stronger than the hydrogen nucleus would, which will make the radius smaller. I'm looking through the equations in this book and I have... [tex]v=\frac{nh}{mr}[/tex] h = h bar [tex]r = n^{2}a_0[/tex] a[SUB]0[/SUB] = Bohr radius [tex]v = \frac{nh}{mn^{2}a_{0}}[/tex] The book doesn't say what r stands for, but it's part of the angular momentum equation L = mrv, so it must be radius of the electron orbit, which is just the radius of the atom. So the radius of the atom is r, but isn't that what the Bohr radius is supposed to be? The formula for a[SUB]0[/SUB] is... [tex]a_{0}=\frac{4∏ε_{0}h^{2}}{me^{2}}[/tex] So I'm guessing the key has to do with the e[SUP]2[/SUP] in the denominator, since the charges of the hydrogen and helium are different. Thanks, was wondering why it wasn't working. [/QUOTE]
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Bohr radius of a helium ion
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