Asked by sweety
Analyze the radial distribution of the H-atom at the 2s state. How are the maxima
situated with respect to the minimum?
Is the “golden ratio” involved?
situated with respect to the minimum?
Is the “golden ratio” involved?
Answers
Answered by
drwls
The 2s orbital has a probability distribution function that is a function of r only.
There is a maximum at r = 0 and a zero-probability node at r = 2 ao, where ao is the Bohr radius. There is a secondary spherical-shell relative maximum at r = 4 ao. In don't see where the "golden ratio" (1.618) is involved.
The probability distribution function of that state is
u^2(r) = [1/(32 pi)]*(1/ao)^3 *[(2 -(r/ao)]^2 *exp(r/ao)
See if you can find a golden ratio in the min and max locations
There is a maximum at r = 0 and a zero-probability node at r = 2 ao, where ao is the Bohr radius. There is a secondary spherical-shell relative maximum at r = 4 ao. In don't see where the "golden ratio" (1.618) is involved.
The probability distribution function of that state is
u^2(r) = [1/(32 pi)]*(1/ao)^3 *[(2 -(r/ao)]^2 *exp(r/ao)
See if you can find a golden ratio in the min and max locations
Answered by
drwls
The probability distribution function of that state is
u^2(r) = [1/(32 pi)]*(1/ao)^3 *[(2 -(r/ao)]^2 *exp(-r/ao)
u^2(r) = [1/(32 pi)]*(1/ao)^3 *[(2 -(r/ao)]^2 *exp(-r/ao)
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