The Structure of the Atom

Section Learning Objectives

By the end of this section, you will be able to do the following:

• Describe Rutherford’s experiment and his model of the atom
• Describe emission and absorption spectra of atoms
• Describe the Bohr model of the atom
• Calculate the energy of electrons when they change energy levels
• Calculate the frequency and wavelength of emitted photons when electrons change energy levels
• Describe the quantum model of the atom

Section Key Terms

energy-level diagram	excited state	Fraunhofer lines
ground state	Heisenberg Uncertainty Principle	hydrogen-like atoms
planetary model of the atom	Rutherford scattering	Rydberg constant

How do we know that atoms are really there if we cannot see them with our own eyes? While often taken for granted, our knowledge of the existence and structure of atoms is the result of centuries of contemplation and experimentation. The earliest known speculation on the atom dates back to the fifth century B.C., when Greek philosophers Leucippus and Democritus contemplated whether a substance could be divided without limit into ever smaller pieces. Since then, scientists such as John Dalton (1766–1844), Amadeo Avogadro (1776–1856), and Dmitri Mendeleev (1834–1907) helped to discover the properties of that fundamental structure of matter. While much could be written about any number of important scientific philosophers, this section will focus on the role played by Ernest Rutherford (1871–1937). Though his understanding of our most elemental matter is rooted in the success of countless prior investigations, his surprising discovery about the interior of the atom is most fundamental in explaining so many well-known phenomena.

Rutherford’s Experiment

In the early 1900’s, the plum pudding model was the accepted model of the atom. Proposed in 1904 by J. J. Thomson, the model suggested that the atom was a spherical ball of positive charge, with negatively charged electrons scattered evenly throughout. In that model, the positive charges made up the pudding, while the electrons acted as isolated plums. During its short life, the model could be used to explain why most particles were neutral, although with an unbalanced number of plums, electrically charged atoms could exist.

When Ernest Rutherford began his gold foil experiment in 1909, it is unlikely that anyone would have expected that the plum pudding model would be challenged. However, using a radioactive source, a thin sheet of gold foil, and a phosphorescent screen, Rutherford would uncover something so great that he would later call it “the most incredible event that has ever happened to me in my life”[James, L. K. (1993). Nobel Laureates in Chemistry, 1901–1992. Washington, DC: American Chemical Society.]

The experiment that Rutherford designed is shown in Figure 22.2. As you can see in, a radioactive source was placed in a lead container with a hole in one side to produce a beam of positively charged helium particles, called alpha particles. Then, a thin gold foil sheet was placed in the beam. When the high-energy alpha particles passed through the gold foil, they were scattered. The scattering was observed from the bright spots they produced when they struck the phosphor screen.

Figure 22.2 Rutherford’s experiment gave direct evidence for the size and mass of the nucleus by scattering alpha particles from a thin gold foil. The scattering of particles suggests that the gold nuclei are very small and contain nearly all of the gold atom’s mass. Particularly significant in showing the size of the nucleus are alpha particles that scatter to very large angles, much like a soccer ball bouncing off a goalie’s head.

The expectation of the plum pudding model was that the high-energy alpha particles would be scattered only slightly by the presence of the gold sheet. Because the energy of the alpha particles was much higher than those typically associated with atoms, the alpha particles should have passed through the thin foil much like a supersonic bowling ball would crash through a few dozen rows of bowling pins. Any deflection was expected to be minor, and due primarily to the electrostatic Coulomb force between the alpha particles and the foil’s interior electric charges.

However, the true result was nothing of the sort. While the majority of alpha particles passed through the foil unobstructed, Rutherford and his collaborators Hans Geiger and Ernest Marsden found that alpha particles occasionally were scattered to large angles, and some even came back in the direction from which they came! The result, called Rutherford scattering, implied that the gold nuclei were actually very small when compared with the size of the gold atom. As shown in Figure 22.3, the dense nucleus is surrounded by mostly empty space of the atom, an idea verified by the fact that only 1 in 8,000 particles was scattered backward.

Figure 22.3 An expanded view of the atoms in the gold foil in Rutherford’s experiment. Circles represent the atoms that are about 10⁻¹⁰ m in diameter, while the dots represent the nuclei that are about 10⁻¹⁵ m in diameter. To be visible, the dots are much larger than scale—if the nuclei were actually the size of the dots, each atom would have a diameter of about five meters! Most alpha particles crash through but are relatively unaffected because of their high energy and the electron’s small mass. Some, however, strike a nucleus and are scattered straight back. A detailed analysis of their interaction gives the size and mass of the nucleus.

Although the results of the experiment were published by his colleagues in 1909, it took Rutherford two years to convince himself of their meaning. Rutherford later wrote: “It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you. On consideration, I realized that this scattering backwards … [meant] … the greatest part of the mass of the atom was concentrated in a tiny nucleus.” In 1911, Rutherford published his analysis together with a proposed model of the atom, which was in part based on Geiger’s work from the previous year. As a result of the paper, the size of the nucleus was determined to be about 10−1510−15 m, or 100,000 times smaller than the atom. That implies a huge density, on the order of 10151015 g/cm³, much greater than any macroscopic matter.

Based on the size and mass of the nucleus revealed by his experiment, as well as the mass of electrons, Rutherford proposed the planetary model of the atom. The planetary model of the atom pictures low-mass electrons orbiting a large-mass nucleus. The sizes of the electron orbits are large compared with the size of the nucleus, and most of the atom is a vacuum. The model is analogous to how low-mass planets in our solar system orbit the large-mass Sun. In the atom, the attractive Coulomb force is analogous to gravitation in the planetary system (see Figure 22.4).

Figure 22.4 Rutherford’s planetary model of the atom incorporates the characteristics of the nucleus, electrons, and the size of the atom. The model was the first to recognize the structure of atoms, in which low-mass electrons orbit a very small, massive nucleus in orbits much larger than the nucleus. The atom is mostly empty and is analogous to our planetary system.

Virtual Physics

Rutherford Scattering

How did Rutherford figure out the structure of the atom without being able to see it? Explore the answer through this simulation of the famous experiment in which he disproved the plum pudding model by observing alpha particles bouncing off atoms and determining that they must have a small core.

Tips For Success

As you progress through the model of the atom, consider the effect that experimentation has on the scientific process. Ask yourself the following: What would our model of the atom be without Rutherford’s gold foil experiment? What further understanding of the atom would not have been gained? How would that affect our current technologies? Though often confusing, experiments taking place today to further understand composition of the atom could perhaps have a similar effect.

Absorption and Emission Spectra

In 1900, Max Planck recognized that all energy radiated from a source is emitted by atoms in quantum states. How would that radical idea relate to the interior of an atom? The answer was first found by investigating the spectrum of light or emission spectrum produced when a gas is highly energized.

Figure 22.5 shows how to isolate the emission spectrum of one such gas. The gas is placed in the discharge tube at the left, where it is energized to the point at which it begins to radiate energy or emit light. The radiated light is channeled by a thin slit and then passed through a diffraction grating, which will separate the light into its constituent wavelengths. The separated light will then strike the photographic film on the right.

The line spectrum shown in part (b) of Figure 22.5 is the output shown on the film for excited iron. Note that this spectrum is not continuous but discrete. In other words, only particular wavelengths are emitted by the iron source. Why would that be the case?

Figure 22.5 Part (a) shows, from left to right, a discharge tube, slit, and diffraction grating producing a line spectrum. Part (b) shows the emission spectrum for iron. The discrete lines imply quantized energy states for the atoms that produce them. The line spectrum for each element is unique, providing a powerful and much-used analytical tool, and many line spectra were well known for many years before they could be explained with physics. (credit:(b) Yttrium91, Wikimedia Commons)

The spectrum of light created by excited iron shows a variety of discrete wavelengths emitted within the visible spectrum. Each element, when excited to the appropriate degree, will create a discrete emission spectrum as in part (b) of Figure 22.5. However, the wavelengths emitted will vary from element to element. The emission spectrum for iron was chosen for Figure 22.5 solely because a substantial portion of its emission spectrum is within the visible spectrum. Figure 22.6 shows the emission spectrum for hydrogen. Note that, while discrete, a large portion of hydrogen emission takes place in the ultraviolet and infrared regions.

Figure 22.6 A schematic of the hydrogen spectrum shows several series named for those who contributed most to their determination. Part of the Balmer series is in the visible spectrum, while the Lyman series is entirely in the ultraviolet, and the Paschen series and others are in the infrared. Values of n_f and n_i are shown for some of the lines. Their importance will be described shortly.

Just as an emission spectrum shows all discrete wavelengths emitted by a gas, an absorption spectrum will show all light that is absorbed by a gas. Black lines exist where the wavelengths are absorbed, with the remainder of the spectrum lit by light is free to pass through. What relationship do you think exists between the black lines of a gas’s absorption spectrum and the colored lines of its emission spectrum? Figure 22.7 shows the absorption spectrum of the Sun. The black lines are called Fraunhofer lines, and they correspond to the wavelengths absorbed by gases in the Sun’s exterior.

Figure 22.7 The absorption spectrum of the Sun. The black lines appear at wavelengths absorbed by the Sun’s gas exterior. The energetic photons emitted from the Sun’s interior are absorbed by gas in its exterior and reemitted in directions away from the observer. That results in dark lines within the absorption spectrum. The lines are called Fraunhofer lines, in honor of the German physicist who discovered them. Lines similar to those are used to determine the chemical composition of stars well outside our solar system.

Bohr’s Explanation of the Hydrogen Spectrum

To tie the unique signatures of emission spectra to the composition of the atom itself would require clever thinking. Niels Bohr (1885–1962), a Danish physicist, did just that, by making immediate use of Rutherford’s planetary model of the atom. Bohr, shown in Figure 22.8, became convinced of its validity and spent part of 1912 at Rutherford’s laboratory. In 1913, after returning to Copenhagen, he began publishing his theory of the simplest atom, hydrogen, based on Rutherford’s planetary model.

Figure 22.8 Niels Bohr, Danish physicist, used the planetary model of the atom to explain the atomic spectrum and size of the hydrogen atom. His many contributions to the development of atomic physics and quantum mechanics, his personal influence on many students and colleagues, and his personal integrity, especially in the face of Nazi oppression, earned him a prominent place in history. (credit: Unknown Author, Wikimedia Commons)

Bohr was able to derive the formula for the hydrogen spectrum using basic physics, the planetary model of the atom, and some very important new conjectures. His first conjecture was that only certain orbits are allowed: In other words, in an atom, the orbits of electrons are quantized. Each quantized orbit has a different distinct energy, and electrons can move to a higher orbit by absorbing energy or drop to a lower orbit by emitting energy. Because of the quantized orbits, the amount of energy emitted or absorbed must also be quantized, producing the discrete spectra seen in Figure 22.5 and Figure 22.7. In equation form, the amount of energy absorbed or emitted can be found as

ΔE=Ei−Ef,ΔE=Ei−Ef,

22.1

where EiEi refers to the energy of the initial quantized orbit, and EfEf refers to the energy of the final orbits. Furthermore, the wavelength emitted can be found using the equation

hf=Ei−Ef,hf=Ei−Ef,

22.2

and relating the wavelength to the frequency found using the equation v=fλv=fλ, where v corresponds to the speed of light.

It makes sense that energy is involved in changing orbits. For example, a burst of energy is required for a satellite to climb to a higher orbit. What is not expected is that atomic orbits should be quantized. Quantization is not observed for satellites or planets, which can have any orbit, given the proper energy (see Figure 22.9).

Figure 22.9 The planetary model of the atom, as modified by Bohr, has the orbits of the electrons quantized. Only certain orbits are allowed, explaining why atomic spectra are discrete or quantized. The energy carried away from an atom by a photon comes from the electron dropping from one allowed orbit to another and is thus quantized. The same is true for atomic absorption of photons.

Figure 22.10 shows an energy-level diagram, a convenient way to display energy states. Each of the horizontal lines corresponds to the energy of an electron in a different orbital. Energy is plotted vertically with the lowest or ground state at the bottom and with excited states above. The vertical arrow downwards shows energy being emitted out of the atom due to an electron dropping from one excited state to another. That would correspond to a line shown on the atom’s emission spectrum. The Lyman series shown in Figure 22.6 results from electrons dropping to the ground state, while the Balmer and Paschen series result to electrons dropping to the n = 2 and n = 3 states, respectively.

Figure 22.10 An energy-level diagram plots energy vertically and is useful in visualizing the energy states of a system and the transitions between them. This diagram is for the hydrogen-atom electrons, showing a transition between two orbits having energies E4E4 and E2E2 . The energy transition results in a Balmer series line in an emission spectrum.

Energy and Wavelength of Emitted Hydrogen Spectra

The energy associated with a particular orbital of a hydrogen atom can be found using the equation

En=−13.6eVn2(n=1,2,3,…),En=−13.6eVn2(n=1,2,3,…),

22.3

where n corresponds to the orbital value from the atom’s nucleus. The negative value in the equation is based upon a baseline energy of zero when the electron is infinitely far from the atom. As a result, the negative value shows that energy is necessary to free the electron from its orbital state. The minimum energy to free the electron is also referred to as its binding energy. The equation is only valid for atoms with single electrons in their orbital shells (like hydrogen). For ionized atoms similar to hydrogen, the following formula may be used.

En=Z2n2Eo(n=1,2,3,…)En=Z2n2Eo(n=1,2,3,…)

22.4

Please note that EoEo corresponds to –13.6 eV, as mentioned earlier. Additionally, ZZ refers to the atomic number of the element studied. The atomic number is the number of protons in the nucleus—it is different for each element. The above equation is derived from some basic physics principles, namely conservation of energy, conservation of angular momentum, Coulomb’s law, and centripetal force. There are three derivations that result in the orbital energy equations, and they are shown below. While you can use the energy equations without understanding the derivations, they will help to remind you of just how valuable those fundamental concepts are.

Derivation 1 (Finding the Radius of an Orbital)

One primary difference between the planetary model of the solar system and the planetary model of the atom is the cause of the circular motion. While gravitation causes the motion of orbiting planets around an interior star, the Coulomb force is responsible for the circular shape of the electron’s orbit. The magnitude of the centripetal force is mev2rnmev2rn, while the magnitude of the Coulomb force is k(Zqe)(qe)re2k(Zqe)(qe)re2 . The assumption here is that the nucleus is more massive than the stationary electron, and the electron orbits about it. That is consistent with the planetary model of the atom. Equating the Coulomb force and the centripetal force,

mev2rn=k(Zqe)(qe)re2,mev2rn=k(Zqe)(qe)re2,

22.5

which yields

rn=k(Zqe2)mv2.rn=k(Zqe2)mv2.

22.6

Derivation 2 (Finding the Velocity of the Orbiting Electron)

Bohr was clever enough to find a way to calculate the electron orbital energies in hydrogen. That was an important first step that has been improved upon, but it is well worth repeating here, because it does correctly describe many characteristics of hydrogen. Assuming circular orbits, Bohr proposed that the angular momentum L of an electron in its orbit is also quantized, that is, it has only specific, discrete values. The value for L is given by the formula

L=mevrn=nh2π(n=1,2,3,…),L=mevrn=nh2π(n=1,2,3,…),

22.7

where L is the angular momentum, m_e is the electron’s mass, r_n is the radius of the n th orbit, and h is Planck’s constant. Note that angular momentum is L=IωL=Iω . For a small object at a radius r, I=mr2I=mr2, and ω=vrω=vr, so that L=Iω=(mr2)(vr)=mvr.L=Iω=(mr2)(vr)=mvr. Quantization says that the value of mvr can only be equal to h / 2, 2h / 2, 3h / 2, etc. At the time, Bohr himself did not know why angular momentum should be quantized, but by using that assumption, he was able to calculate the energies in the hydrogen spectrum, something no one else had done at the time.

Derivation 3 (Finding the Energy of the Orbiting Electron)

To get the electron orbital energies, we start by noting that the electron energy is the sum of its kinetic and potential energy.

En=KE+PEEn=KE+PE

22.8

Kinetic energy is the familiar KE=12mv2KE=12mv2, assuming the electron is not moving at a relativistic speed. Potential energy for the electron is electrical, or PE=qeVPE=qeV, where V is the potential due to the nucleus, which looks like a point charge. The nucleus has a positive charge ZqeZqe; thus, V=kZqernV=kZqern, recalling an earlier equation for the potential due to a point charge from the chapter on Electricity and Magnetism. Since the electron’s charge is negative, we see that PE=−kZqe2rn.PE=−kZqe2rn. Substituting the expressions for KE and PE,

En=12mev2−kZqe2rn.En=12mev2−kZqe2rn.

22.9

Now we solve for r_n and v using the equation for angular momentum L=mevrn=nh2π(n=1,2,3,…)L=mevrn=nh2π(n=1,2,3,…), giving

v=nh2πmern(n=1,2,3,…)v=nh2πmern(n=1,2,3,…)

22.10

and

rn=nh2πmev(n=1,2,3,…).rn=nh2πmev(n=1,2,3,…).

22.11

Substituting the expression for r_n and v into the above expressions for energy (KE and PE), and performing algebraic manipulation, yields

En=−Z2n2Eo(n=1,2,3,…)En=−Z2n2Eo(n=1,2,3,…)

22.12

for the orbital energies of hydrogen-like atoms. Here, E_o is the ground-state energy (n = 1) for hydrogen (Z = 1) and is given by

Eo=2π2qe4mek2h2=13.6eV.Eo=2π2qe4mek2h2=13.6eV.

22.13

Thus, for hydrogen,

En=−13.6eVn2(n=1,2,3,…).En=−13.6eVn2(n=1,2,3,…).

22.14

The relationship between orbital energies and orbital states for the hydrogen atom can be seen in Figure 22.11.

Figure 22.11 Energy-level diagram for hydrogen showing the Lyman, Balmer, and Paschen series of transitions. The orbital energies are calculated using the above equation, first derived by Bohr.