The Chemical Bond
http://www.catalysis.nl/~chembond/

Course Information
Course contents
Lecture Notes
Exercises
Contacting Information
What's New?
Homepage

The radial probability distribution.

Probability calculations and statistics are littered with paradoxes. As quantum mechanics makes only probabilistic statements, we will also get such paradoxes in quantum mechanics and in the theory of the chemical bond. Suppose we have a hydrogen atom in its ground state. We will deal here with the paradox that the most likely position to find the electron is at the position of the nucleus, and the most likely distance to find the electron is the Bohr radius a0 = 0.529177Å.

According to the Copenhagen interpretation the probability distribution for the electron in the hydrogen atom is given by |Y|2, where Y is the wave function of the electron. The ground state is the 1s function, or

Y1,0,0(r,q,j) = æ
ç
è 		1
pa03
 ö
÷
ø 		1/2

 
 exp é
ê
ë 		- r
a0
 ù
ú
û 		.
The probability distribution is then
|Y1,0,0(r,q,j)|2 = 1
pa03
 exp é
ê
ë 		- 2r
a0
 ù
ú
û 		.
This probability distribution has a maximum at r = 0, so the most likely place to find the electron is at the nucleus.

Because the 1s wave function depends only on the distance, one might think that |Y1,0,0|2 is also the probability distribution that the electron is found at a certain distance from the nucleus. This is, however, not the case. We need to be precise when we talk about this probability distribution, which is called the radial probability distribution . We write the probability to find the electron at a distance between r and r+dr from the nucleus as Prad(r) dr, where Prad is the radial probability distribution, and dr is an (infinitisimal) small number. This probability is equal to the probability to find the electron at a distance less than r+dr minus the probability to find the electron at a distance less than r. The probability to find the electron at a distance less than r is given by
ó
õ 		r

0 
 dr¢ ó
õ 		p

0 
 dq ó
õ 		2p

0 
 dj 
|Y1,0,0(r¢,q,j)|2(r¢)2sinq
= 1- é
ê
ë 		2 æ
ç
è 		r
a0
 ö
÷
ø 		2

 
 +2 æ
ç
è 		r
a0
 ö
÷
ø 		+1 ù
ú
û
×exp é
ê
ë 		- 2r
a0
 ù
ú
û 		.
The integral is over the sphere with radius r, and the integration is done in spherical coordinates. Because of this we have to use the volume element (r¢)2sinq. The probability to find the electron at a distance less than r+dr can be obtained by making the substitution r® r+dr. To get Prad we take the difference between the result for r+dr and r. We also take dr to be small so that we can neglect quadratic and higher order terms. This gives us
Prad(r) = 4r2
a03
 exp é
ê
ë 		- 2r
a0
 ù
ú
û 		.
If we compare this with |Y1,0,0|2 we see that
Prad = 4pr2|Y1,0,0|2.
We see that there is a factor 4pr2. This factor is responsible for the fact that the radial probability distribution is zero at r = 0 whereas the wave function has a maximum at r = 0. The next figure shows the origin of this factor.

The probability to find the electron at a certain distance from the nucleus is not equal to the probability to find the electron at a certain point at that distance from the nucleus. There is a whole surface of a sphere on which we can find the electron at that distance. The area of that surface equals 4pr2. The radial probability distribution has a maximum at r = a0. We are referring to this maximum when we say that the electron in the 1s wave function is most likely to be found at a distance r = a0 from the nucleus.
Last updated: Thursday, March 20, 2003
© Dr. A.P.J. Jansen