Finding maximum of a wavefunction?

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smileandbehappy
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Homework Statement



http://[ATTACH=full]199743[/ATTACH] [ATTACH=full]199744[/ATTACH]

Question D.

[h2]Homework Equations[/h2]

[ATTACH=full]199745[/ATTACH]

[h2]The Attempt at a Solution[/h2]

The wavefunction is all real. So I can simply sqaure it... However when I do this and differentiate it using the product rule, I'm getting r = 2a(naught) as the maximum and not 4a(naught)... Any advice?

[ATTACH=full]199746[/ATTACH]
 
Attachments
  • Textbook problem showing the normalized hydrogen wavefunction psi(2,1,-1) = (1/(64 a0^3))^(1/2) (r/a0) e^(-r/2a0) sin(theta) e^(-i phi), followed by parts (a) to (e) of the question.
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  • Typeset exam question showing the hydrogen atom eigenfunction psi(2,1,-1)(r, theta, phi) = (1/(64 pi a0^3))^(1/2) (r/a0) e^(-r/2a0) sin(theta) e^(-i phi), with definitions of quantum numbers and parts (a) to (e) below.
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  • Table 2.1 of normalized radial functions R_nl(r) for the hydrogen atom, listing n, l and the expressions for n = 1 to 3, above plots of R_nl(r)/a0^(-3/2) against r/a0 for n = 1, 2 and 3 at various l values.
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  • Handwritten derivation on lined notebook paper finding the maximum of |R21|^2 by differentiating A r^2 e^(-r/a0) with the product rule, giving dR/dr = A r e^(-r/a0) [2 - r/a0], then setting it to zero and substituting r = 4a0 with scribbled working.
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Last edited by a moderator:
The probability of finding the electron is some volume element dV is |ψ|2 dV. The dV is important.

In spherical coordinates, dV is proportional to r2.
 
Thanks - feel a little foolish now... Here is my updated workings incase it's useful for someone in the future:

Handwritten derivation finding the radial maximum of the hydrogen 2p radial probability density, starting from P = R^2 r^2 and ending with the turning point at r = 4a0.


Please let me know if I've made any horrific mistakes.