Physical Chemistry , 1st ed.

index nis the quantum number,and it can have any integer value from 0 to in-

finity. (As we will see from the form of the wavefunction, 0 is a possible value

for the quantum number in this case.)

Before returning to wavefunctions, we have a few points to consider with re-

spect to the total energy. A diagram of the energy levels for different quantum

numbers (assuming the mass and the force constant remain the same) is shown

in Figure 11.2. For an ideal harmonic oscillator, the energy levels are spaced

by the same amount. It is easy to show that the energy levels are separated by

Eh. Further, the lowest possible value for energy is not zero. This is seen

by substituting the lowest possible value for the quantum number n, which is

zero. We get

E(n0) (0 ^12 )h

^12 h

which is a nonzero value for the total energy. This introduces the concept of

zero-point energy.At the minimum value for the quantum number (the ground

stateof the oscillator), there is still a nonzeroamount of energy in the system.

The frequency, , should be in units of s^1. Multiplying s^1 by the units on

Planck’s constant, J s or erg s, results in units of J or erg, which are units of en-

ergy. It is common to express the energy difference in terms of the photon used

to excite a system from one energy level to another, since harmonic oscillators

can go from state to state by the absorption or emission of a photon, just as

with Bohr’s hydrogen atom. One characteristic used to describe the photon is

its wavelength. Using the equation c

(where cis the speed of light and 

is its wavelength), one can convert from wavelength to frequency. The follow-

ing examples illustrate.

Example 11.3

A single oxygen atom attached to a smooth metal surface vibrates at a fre-

quency of 1.800 

1013 s^1. Calculate its total energy for the n0, 1, and 2

quantum numbers.

Solution:

We use equation 11.16 with 

1.800 

1013 s^1 and n0, 1, and 2:

E(n0) (0  2 ^1 )(6.626 

10 ^34 J s)(1.800 

1013 s^1 )

E(n1) (1  2 ^1 )(6.626 

10 ^34 J s)(1.800 

1013 s^1 )

E(n2) (2  2 ^1 )(6.626 

10 ^34 J s)(1.800 

1013 s^1 )

From the above expressions, we get

E(n0) 5.963 

10 ^21 J

E(n1) 1.789 

10 ^20 J

E(n2) 2.982 

10 ^20 J

The minimum energy of this vibrating oxygen atom, its zero-point energy, is

5.963 

10 ^21 J.

Example 11.4

Calculate the wavelength of light necessary to excite a harmonic oscillator

from one energy state to the adjacent higher state in Example 11.3. Express

the wavelength in units of m,m (micrometers), and Å.

11.3 The Quantum-Mechanical Harmonic Oscillator 323

n  4 E ^92 h

n  3 E ^72 h

x

Energy

n  2 E ^52 h

n  1 E ^32 h

n  0 E ^12 h

Figure 11.2 A diagram of the energy levels of
an ideal harmonic oscillator, as predicted by the
solutions to the Schrödinger equation. Note that
the lowest quantized level,E(n0), does not
have zero energy.