... why are there more nodes/zeros?(adsbygoogle = window.adsbygoogle || []).push({});

If l = 0 then the angular momentum contribution to the effective potential is zero, and there is the coublomb potential only. So shouldn't it always go as ~ -k/r^2 (k = constant) like the n=1 s orbital?

Why is it that for n = 2 is there 1 zero, n= 3 there is 2 zeros, etc?

Thanks

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# L = 0 (s) orbital potentials V(r) as n increases - why are..

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