PhysicsLAB: Energy-Level Diagrams

Energy level diagrams are a means of analyzing the energies electrons can accept and release as they transition from one accepted orbital to another. These energies differences correspond to the wavelengths of light in the discreet spectral lines emitted by an atom as it goes through de-excitation or by the wavelengths absorbed in an absorption spectrum.

Using the Bohr Model, the energy levels (in electron volts, eV) are calculated with the formula:

E_n = -13.6 (Z²/n²) eV

where Z is the atomic number and n is the energy level. The ground state is represented by n = 1, first excited state by n = 2, second excited state by n = 3, etc.

The unit eV means an electron volt and represents an easy way to state the kinetic energy gained by a charged particle when it is accelerated across an electric potential difference.

KE = q(ΔV)

The charge on an electron is 1e or 1.6 x 10^-19 coulombs. When it is accelerated across a potential of 1 volt (1 J/C) then the kinetic energy it gains equals

ΔKE = 1 eV = (1.6 x 10^-19 C)(1 J/C) = 1.6 x 10^-19 J

To use most formulas, energies, when given in eV, must first be converted into Joules.

Using Bohr's formula, a hypothetical, doubly-ionized atom with Z = 3 could have the following energy level diagram.

Notice how each energy level closer and closer to the nucleus is more and more negative. This signifies that the electron is trapped in an "energy well." To ionize a ground-state electron [to take it from -122.4 eV to 0 eV in our example], you would have to irradiate the gas with photons having energies of 122.4 eV or greater. This is the ONLY instance where the incident energy does not have to EXACTLY match the difference in two energy levels. Any excess energy would remain in the form of the ionized electron's kinetic energy.

Max Planck had already determined that the energy levels in an oscillating system were quantized and followed the relationship

E = hf

where f represented the frequency of oscillation and h is Planck's constant, 6.63 x 10^-34 J sec. In our case, f is the frequency of the emitted photon in accordance with c = fλ. If we replace f with c/λ, we have the formula

E = h(c/λ)

The value of E, or energy, represents the difference in the energies of two energy levels (ΔE) when an electron goes through de-excitation. The value of λ corresponds to the wavelength of the emitted spectral line.

λ = hc/ΔE

Refer to the following information for the next three questions.

Suppose an electron of our hypothetical atom has been excited to its second excited state (n = 3). When it falls back to its ground state it can do so through a total of three possible transitions. How much energy is released during each of these transitions?

	n = 3 to n = 1

	n = 3 to n = 2

	n = 2 to n = 1

Eight years before Bohr's theory was announced, a pattern in the frequencies of spectral lines was discovered by W. Ritz. The Ritz Combination Principle states that the spectral lines of the elements have frequencies that are either the sums or the differences of the frequencies of two other lines.

Let's use our example to illustrate how Bohr's atomic model supports this addition pattern of the frequency of emission lines. The energies lost, or liberated, by the electron are the energies present in each emitted photon.

electron: -108.8 eV = (-17 eV) + (-91.8 eV)

E_2®ground = E_2®1+ E_1®ground

photon: 108.8 eV = 17 eV + 91.8 eV

hf_2®ground = hf_2®1+ hf_1®ground

f_2®ground = f_2®1+ f_1®ground

Refer to the following information for the next three questions.

Converting our energy values from eV to Joules and using the formula f = E/h, what would be the frequency of each transition we previously calculated?

	f_2®ground

	f_2®1

	f_1®ground

But our concern is emitted wavelengths, not the frequencies.

Refer to the following information for the next three questions.

Convert each frequency into its photon's wavelength. Use the formula l = c/f.

	l _2®ground

	l _2®1

	l _1®ground

All three of these emissions are in the ultraviolet portion of the electromagnetic spectrum, none would be visible. Notice that although the frequencies added together correctly, the wavelengths do NOT.

Type of radiation

Range of wavelengths

radio

570 down to 2.8 meters

5.6 down to 0.34 meters

microwave

0.1 down to 0.001 meters

infrared radiation

10^-3 down to 10^-7 meters

visible light


red
orange
yellow
green
blue
indigo
violet

700 to 400 nm

ultraviolet

10^-7 down to 10^-10 meters

x-rays

10^-10 down to 10^-12 meters

gamma rays

shorter than 10^-12 meters

Refer to the following information for the next question.

Suppose that we know that an electron emits a photon having a wavelength of 1.43 x 10^-7 meters when it transitions from an unknown energy level to the second excited state.

	On which energy level did it initially start?