Physical Chemistry , 1st ed.

where ais the same constant previously defined for the Rfunctions. Since it is
defined as a group of constants,aitself is a constant and has units of length
(shown in equation 11.69 in units of Å). The constant ais called the Bohr ra-
dius.This most probable distance is exactly the same distancethat an electron
of Bohr’s theory would have in its first orbit. Quantum mechanics does not
constrain the distance of the electron from the nucleus as did Bohr’s theory.
But it does predict that the distance Bohr calculated for the electron in its
lowest energy state is in fact the most probabledistance of the electron from
the nucleus. (It is sometimes written a 0 , which is defined similarly but uses the
mass of the electron instead of the reduced mass of the hydrogen atom. The
difference is very slight.)

Example 11.24

a.What is the probability that an electron in the  1 sorbital of hydrogen will

be within a radius of 2.00 Å from the nucleus?

b.Calculate a similar probability, but now for an electron within 0.250 Å of

a Be^3 nucleus.

Solution

a.For a normalized wavefunction, the probability Pis equal to

P

b

a

*d

where aand bare the limits of the space being considered. For the hydrogen

atom, this becomes the three-dimensional expression

P

a^3

1





2 

0

d



0

sin d 

2.00 Å

0

r^2 e^2 r/adr

where the wavefunction in terms of the Bohr radius ahas already been

squared and the expression has been separated into three integrals. The two

angular integrals we have done before, and the integral over rcan be found

in Appendix 1. The expression becomes

P

a^3

1



 2  (^2) e^2 r/a



2

r^2 a



r

2

a^2



a

4

3

 0 2.00Å

If the value ofain units of angstroms, 0.529 Å, is used in the above expres-

sion, then the 2.00-Å limit can be used directly because the quantities are ex-

pressed in the same units. Substituting and evaluating the expression at the

limits:

P

(0.529

1

Å)^3 

 2 2[(5.201 

10 ^4 )(1.337841)Å^3 

(1)(3.701 

10 ^2 )Å^3 ]

Note that the Å^3 units cancel from the expression, and the probability is unit-

less (as it should be). Evaluating this expression, we find that

P0.981, or 98.1%

This example shows that the electron has a 98.1% probability of being within

2.00 Å, or slightly under 4 Bohr radii, from the nucleus. You might want to

compare this with Figure 11.19, where the total probability is represented by

the area under the curve. Finally, note that this implies a 1.9% chance that

the electron is fartherthan 2.00 Å from the nucleus.

11.11 The Hydrogen Atom Wavefunctions 361