Physical Chemistry Third Edition

738 17 The Electronic States of Atoms. I. The Hydrogen Atom

From the first relation in Eq. (17.3-4), the energy is quantized and determined by the

value ofn:

EEn− ̄

h^2 α^2

2 μ

−

μe^4

2(4πε 0 hn ̄ )^2

−

2 π^2 μe^4

(4πε 0 h)^2

1

n^2

(17.3-9)

This is the same expression for the energy as in the Bohr theory. However, the quan-

tization now arises from the Schrödinger equation and its boundary conditions, and

not from an arbitrary assumption as in the Bohr theory. Since the Bohr theory is based

on assuming the wrong values for the angular momentum, its success with the energy

eigenvalues seems to be fortuitous and at the least is somewhat mysterious.

The zero of the potential energy function is chosen so that it vanishes atr→∞. The

negative values of the energy eigenvalueEin Eq. (17.3-9) correspond tobound states,

in which there is not enough energy for the electron to escape from the nucleus. There

are also nonbound states calledscattering statesin which the energy is positive so that

the electron moves toward the nucleus, passes it, and continues on its way, generally in

a new direction. We will not discuss these states, which do not have quantized energy

values.^3

Exercise 17.6

Substitute the values of the constants into Eq. (17.3-9) to show that the relative energy of a

hydrogen atom can take on the values

EEn−

2. 1787 × 10 −^18 J

n^2

−

13 .60 eV

n^2

(17.3-10)

where 1 eV (1electron volt) is the energy required to move one electron through an electric

potential difference of 1 volt, equal to 1. 602177 × 10 −^19 J.

TheBohr radiusgiven in Eq. (14.4-15) is equal to the radius of the first orbit of a

hydrogen atom in the Bohr theory assuming a stationary nucleus. If we correct for the

motion of the nucleus by replacing the electron mass by the reduced mass,

a

h ̄^24 πε 0

μ^2

 5. 2947 × 10 −^11 m 52 .947 pm 0 .52947 Å (17.3-11)

where Å represents the angstrom unit, 10−^10 m. When we express the energy in terms

of the Bohr radius, we get

EEn−

h ̄^2 α^2

2 μ

−

e^2

2(4πε 0 )an^2

(17.3-12)

This energy is equal to half of the potential energy of an electron at a distance from the

nucleus equal toan^2 , the radius of the corresponding Bohr orbit.

Exercise 17.7

Verify Eqs. (17.3-11) and (17.3-12).

(^3) H. A. Bethe and E. E. Salpeter,Quantum Mechanics of One- and Two-Electron Systems, Plenum,
New York, 1977, p. 21ff, p. 32ff.