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Don't understand these correct answers.
For #8, it seems like each hump of the radial probability distribution represents an S orbital, and so it makes sense for the largest probability to be 4s. But that contradicts question 7, because the correct answer has the Hydrogen electron residing in the 4p. orbital which makes absolutely zero sense to me.
Can anyone hjelp clear this up? Thank you
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Just plug in.
[math]{\Psi _{n\ell m}}\left( {r,\theta ,\varphi } \right) = \sqrt {{{\left( {\frac{{2r}}{{na}}} \right)}^3}\frac{{\left( {n - \ell - 1} \right)!}}{{2n\left( {n + \ell } \right)!}}} {e^{ - r/na}}{\left( {\frac{{2r}}{{na}}} \right)^\ell }L_{n - \ell - 1}^{2\ell + 1}\left( {2r/na} \right)P_\ell ^m\left( {\cos \theta } \right){e^{im\varphi }}[/math]
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>>8548892
That doesn't help!
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>>8548892
>Time independent
>Non-relativistic
>Not taking spin into account
Fucking brainlets
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>>8548890
>each hump of the radial probability distribution represents an S orbital
no, the 4s all by itself has 4 humps (ie it has 3 nodes)
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>>8548890
Ground state is lowest energy. Final state is highest.
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