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Topic: Radial probability distribution (rpd)  (Read 4995 times)

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Offline Shah

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Radial probability distribution (rpd)
« on: March 15, 2012, 10:47:59 AM »
Radial Probability Density = R(r) : Square of the Radial Wavefunction
The required volume is determined by the volume of the SPHERICAL SHELL enclosed between a sphere of radius (r+dr) and a sphere of radius r

rpd = radial probability density × volume of the spherical shell = R2 × 4πr2 dr


how then did they cancel the dr and directly write:
rpd = R2 × 4πr2

??

also how could only radial probability be multiplied by volume? shouldn't the whole wave function be multiplied by the volume

Offline Shah

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Re: Radial probability distribution (rpd)
« Reply #1 on: March 17, 2012, 02:08:51 PM »
?

Offline juanrga

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Re: Radial probability distribution (rpd)
« Reply #2 on: March 20, 2012, 07:23:08 AM »
Radial Probability Density = R(r) : Square of the Radial Wavefunction
The required volume is determined by the volume of the SPHERICAL SHELL enclosed between a sphere of radius (r+dr) and a sphere of radius r

rpd = radial probability density × volume of the spherical shell = R2 × 4?r2 dr


how then did they cancel the dr and directly write:
rpd = R2 × 4?r2

??

also how could only radial probability be multiplied by volume? shouldn't the whole wave function be multiplied by the volume

The probability to find the electron in a region of space between ##r## and ##r + \mathrm{d}r##, ##\theta## and ##\theta + \mathrm{d}\theta## and ##\phi## and ##\phi + \mathrm{d}\phi## is
$$ |\psi|^2 \mathrm{d}\tau = [R_{nl}(r)]^2 |Y_l^m(\theta,\phi)|^2 4\pi r^2 \sin \theta  \mathrm{d}r \mathrm{d}\theta \mathrm{d}\phi$$

where ##R_{nl}## is the radial function and ##Y_l^m## the spherical harmonics.

If you want to know the probability to find the electron in a region of space between ##r## and ##r + \mathrm{d}r## with independence of the other coordinates, then you need to integrate the above expression over all possible values of ##\theta## and ##\phi##. The result is
$$ [R_{nl}(r)]^2 4\pi r^2 \mathrm{d}r \int_0^{2\pi} \int_0^\pi |Y_l^m(\theta,\phi)|^2  \sin \theta \mathrm{d}\theta \mathrm{d}\phi = [R_{nl}(r)]^2 4\pi r^2 \mathrm{d}r$$
Therefore, you have a density × spherical element of volume.

The whole wavefunction density ##|\psi|^2## is multiplied by the whole element of volume ##\mathrm{d}\tau##. For a spherical element of volume ##4\pi r^2 \mathrm{d}r## , only the radial part of the wavefunction density ##[R_{nl}(r)]^2## multiplies this volume.

For your other question, the difference between writing the ## \mathrm{d}r ## or not depends if you are studying the density of probability at point ##r## or the probability between ##r## and ##r + \mathrm{d}r##.
« Last Edit: April 03, 2012, 09:09:35 AM by Borek »
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