40.3.4 Solve II: Find the Allowed Energies

Equation 40.16 defined β. The boundary condition of Equation 40.19 then placed restrictions on the possible values of β:

βn=2mEnℏ=nπL  n=1, 2, 3,  … (40.21)

where the value of β and the energy associated with the integer n have been labeled βn and En. We can solve for En by squaring both sides:

En=n2π2ℏ22mL2=n2h28mL2  n=1, 2, 3,  … (40.22)

where, in the last step, we used the definition ℏ=h/2π. For a particle in a box, these energies are the only values of E for which there are physically meaningful solutions to the Schrödinger equation. That is, the particle’s energy is quantized! It is worth emphasizing that quantization is not inherent in the wave function itself but arises because the boundary conditions—the physics of the situation—are satisfied by only a small subset of the mathematical solutions to the Schrödinger equation.

It is useful to write the energies of the stationary states as

En=n2E1 (40.23)

where En is the energy of the stationary state with quantum number n. The smallest possible energy E1=h2/8mL2 is the energy of the n=1 ground state. These allowed energies are shown in the energy-level diagram of FIGURE 40.6. Recall, from Chapter 38, that an energy-level diagram is not a graph (the horizontal axis doesn’t represent anything) but a “ladder” of allowed energies.

FIGURE 40.6 The energy-level diagram for a particle in a box.

The figure illustrates the energy-level diagram for a particle in a box.
FIGURE 40.6 Full Alternative Text

Equation 40.22 is identical to the energies we found in Chapter 38 by requiring the de Broglie wave of a particle in a box to form a standing wave. Only now we have a theory that tells not only the energies but also the wave functions.

EXAMPLE 40.1 An electron in a box

An electron is confined to a rigid box. What is the length of the box if the energy difference between the first and second states is 3.0 eV?

MODEL Model the electron as a particle in a rigid one-dimensional box of length L.

SOLVE The first two quantum states, with n=1 and n=2, have energies E1 and E2=4E1. Thus the energy difference between the states is

ΔE=3E1=3h28mL2=3.0 eV=4.8×10−19 J

The length of the box for which ΔE=3.0 eV is

L=3h28m ΔE=6.14×10−10 m=0.614 nm

REVIEW The expression for E1 is in SI units, so energies must be in J, not eV.