Physical Chemistry Third Edition

17.3 The Radial Factor in the Hydrogen Atom Wave Function. The Energy Levels of the Hydrogen Atom 739

In the (fictitious) limit that the nucleus is infinitely massive compared to the electron,

the electron moves about the stationary nucleus, and the reduced mass becomes

lim

mn→∞

μ lim

mn→∞

(

memn

me+mn

)



memn

mn

me (17.3-13)

whereme is the mass of the electron. The Bohr radius becomes the same as in

Eq. (14.4-15);

lim

mn→∞

aa 0 

h ̄^24 πε 0

mee^2

 5. 29198 × 10 −^11 m (17.3-14)

Exercise 17.8

Calculate the percentage error in the hydrogen atom Bohr radius and in the hydrogen atom energy

introduced by replacing the reduced mass by the mass of the electron.

The rules that the hydrogen atom quantum numbers obey can be restated:

n1, 2, 3,... (17.3-15a)

l0, 1, 2,...,n− 1 (17.3-15b)

m0,±1,±2,...,±l (17.3-15c)

The quantum numbernis called theprincipal quantum number. It determines the

energy. The quantum numberlhas been called theazimuthal quantum number, but

could also be called theangular momentum quantum number. It determines the magni-

tude of the angular momentum. The quantum numbermhas been called themagnetic

quantum number, but could also be called theangular momentum projection quan-

tum number. It determines thezcomponent of the angular momentum. The energy

eigenvalue depends only on the value of the principal quantum numbern, so the set of

states with a given value ofnbut with different values oflandmconstitute an energy

level that is degenerate. Figure 17.5 depicts the energy levels for the first few bound

states of a hydrogen atom. Each state is represented by a horizontal line segment at the

appropriate height for its energy level.

More states not shown

Continuum of states

2 s 2 p

E/

eV

3 s 3 p 3 d

1 s

4 s 4 p 4 d 4 f

0

25

210

215

Figure 17.5 Energy Levels of the
Hydrogen Atom.

EXAMPLE17.3

Find an expression for the degeneracy of the hydrogen atom energy levels.

Solution

For a given value ofn, the possible values oflrange from 0 ton−1. For a given value of

l, the values ofmrange from−ltol. The number of possible values ofmfor a given value

oflis 2l+1, sincemcan have any oflpositive values, any oflnegative values, or can be

equal to zero. The degeneracygnis

gn

n∑− 1

l 0

(2l+1) 2

n∑− 1

l 0

l+

n∑− 1

l 0

1  2 n

(

0 +n− 1

2

)

+n

n^2 −n+nn^2 (17.3-16)