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21.1The structure of the atom

In the Rutherford experiment, positively charged particles, called alpha particles, were directed at a thin gold foil in an evacuated chamber, Figure 21.1. The number of particles deflected at a particular angle was recorded. They found that:

●the great majority of the alpha particles went straight through the foil with no or little deflection; they were detected at very small scattering angles, such as at positions A, B and C in Figure 21.1a.

●very occasionally, alpha particles were detected at very large scattering angles, such as position L.

Figure 21.1: a The majority of alpha particles are slightly deflected by the gold foil. Very occasionally, large-angle scatterings take place. b The alpha particles are detected by the sparks of light they create when they hit a zinc sulfide screen in the microscope.

The small deflections could be understood within the Thomson model: they were due to the electric force of repulsion between the positive charge of the gold atoms and the positive charge of the alpha particles. (An alpha particle is about 7000 times more massive than the electron, so the effect of the electrons of the gold atoms on the path of the alpha particles is negligible.)

But the large-angle scattering events could not be understood within the Thomson model. Rutherford said that ‘it was as if you fired a 15-inch shell at tissue paper and it came back and hit you’.

Consequences of the Rutherford experiment

The very large deflection showed that there was an enormous force of repulsion between the alpha particle and the positive charge of the atom. Since the electric force is given by F=k q 1 q 2 r 2 to get a large force implies that the separation r must be very small.

How can we get a large force in the Thomson model? Figure 21.2a shows two alpha particle paths. Suppose the alpha particle approaches along the top path. The closest distance to the centre of the positive charge is the atomic radius and this is of order 10−10 m. Calculations show that the resulting force is 1010 times too small, so this does not work. What if the alpha penetrates the atom? Then, the distance r can become as small as we like. But this does not work either. In this case, only the charge within the smaller dotted sphere would produce a force on the alpha particle, and the result is again a very small force and a small deflection. The only way out is to imagine that the positive charge on the atom is within a sphere that is much smaller than the sphere that Thomson had imagined.

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Figure 21.2: a We cannot get a large force out of the Thomson model, but b with a tiny nucleus we can.

Rutherford calculated theoretically the number of alpha particles expected at particular scattering angles based on Coulomb’s force law. He found agreement with his experiments if the positive atomic charge was confined to a region of linear size approximately equal to 10−15 m. This and subsequent experiments confirmed the existence of the nucleus inside the atom—a small, massive object carrying the positive charge of the atom and most of its mass.

Worked Example 21.1

A sphere of charge Q has radius 10−15 m. Another sphere has the same charge and a radius of 10−10 m. Calculate the ratio of the electric fields at the surface of the two spheres.

Answer

Apply the formula for the electric field:

E=k Q r 2

We find the ratio of the fields E 1 and E 2 for the two spheres as:

E 1 E 2=k Q(10−15)2 k Q(10−10)2

E 1 E 2=10 10

This is why the deflecting forces in Rutherford’s nuclear model are so large compared with what one might expect from Thomson’s model.

Exam tip

You must be prepared to explain why the old model can account for the small deflections but not the large deflections, and why the proposed new model by Rutherford explains both the small as well as the large deflections.

CHECK YOURSELF 1

Which feature of the Rutherford experiment led to the idea of the existence of a tiny nucleus?

KEY POINT

The Rutherford experiment established the Rutherford model of the atom: the atom consists of a very small, massive, positively charged nucleus, and the electrons orbit the nucleus like planets around the sun.

Discrete energy

If you expose a container of gas at low pressure to a strong electric field, light will be emitted from the gas. You can show that the emitted light contains many different wavelengths by passing it through a prism or diffraction grating. The result is a series of bands of light at different wavelengths. Figure 21.3 shows the wavelengths that are present in the light emitted by hydrogen, helium and mercury vapour.

The set of possible wavelengths that can be emitted is called the emission spectrum the set of wavelengths of light emitted by an atom.

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Figure 21.3: The emission spectra of hydrogen, helium and mercury vapour.

How can these spectra be understood? Niels Bohr (1885–1962) provided the first radical explanation in 1913. He argued that an atom has discrete energy energy taking a set of specific values rather than a continuous set of values, i.e. it can have one out of a specific set of values. He represented the possible energies with an energy level diagram a representation of the discrete energies of an atom. Each horizontal level represents a possible energy of the atom. By ‘energy of the atom’ we mean the kinetic energy of the electrons plus the electrical potential energy of the electrons and the nucleus. The diagrams for hydrogen and mercury are shown in Figure 21.4.

Bohr applied his theory to hydrogen, the simplest atom with just one electron. He was able to show that the energy of the hydrogen atom in the nth state is given by:

E=−13.6 n 2 eV n=1,2,3,…

Exam tip

This formula only applies to hydrogen.

Figure 21.4: Energy level diagrams according to Bohr: a for hydrogen, and b for mercury.

The negative sign in the energy tells us that the electron is trapped within the atom. To have the electron leave the atom, energy must be provided. A hydrogen atom can only have an energy of −13.6 eV, −3.40 eV, −1.51 eV, −0.87 eV and so on. No other value is possible. Energy in the atomic world is discrete.

CHECK YOURSELF 2

How much energy must be given to a hydrogen atom in its ground state (the lowest energy state) so that the electron is knocked out of the atom? This is known as the ionisation energy.

How does this energy level structure help explain emission spectra? Bohr suggested that an atom can make a transition from a state of higher energy to a state of lower energy by emitting a photon the particle of light (a quantum of energy), a massless particle moving at the speed of light, the particle of light.

The energy of the emitted photon is the difference in energy between the two levels. Think of the photon as a ‘tiny flash of light’. There would be one photon for each transition. With very many transitions from very many atoms the ‘tiny flashes of light’ in each transition add up to the observable light we see in the emission spectrum. The photon had been introduced earlier into physics by Einstein, who suggested that its energy is given by:

E=h f or E=h c λ

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where f and λ are the frequency and wavelength of the light (the photon), c is the speed of light and h a new constant introduced by Planck a few years earlier. The Planck constant has a value 6.63 × 10−34 J s.

In a transition from a high to lower energy state, such that the difference in energy between the two states is ΔE, the photon emitted has a wavelength given by

h c λ=Δ E and therefore

λ=h c Δ E

As an example, consider the transition from the level n = 3 (whose energy is −1.51 eV) to the level n = 2 (whose energy is −3.40 eV). The difference in energy between these two levels is −1.51 − (−3.40) = 1.89 eV, and this is the energy that will be carried by the photon emitted in this transition. Therefore:

h c λ=1.89 eV=1.89×1.6×10−19=3.024×10−19 J λ=6.63×10−34×3.0×10 8 3.024×10−19 λ=6.58×10−7 m

Exam tip

Remember to convert eV into joules!

This is in excellent agreement with the red line in the spectrum of hydrogen in Figure 21.3.

In hydrogen, most transitions from higher levels to the level n = 2 emit photons of visible light.

When undisturbed, the electron in each hydrogen atom will occupy the lowest energy state, i.e. the one with n = 1 and energy −13.6 eV. This lowest energy state is called the ground state the state of lowest energy. If energy is supplied to the atom, the electron may move to a higher energy level, an excited state any state above the ground state, by absorbing exactly the right amount of energy needed to move up. For example, to move from n = 1 to the state with n = 3 the energy needed is exactly 13.6 − 1.51 = 12.09 eV. Suppose that precisely this amount of energy is supplied to an electron in the ground state. The electron will absorb this energy and make a transition to the level n = 3. At this point the atom is said to be excited.

From the excited state, the electron will immediately (within nanoseconds) make a transition down to one of the available lower energy states. From n = 3 the electron can either go directly to n = 1 (emitting a photon of energy 12.1 eV) or it can first make a transition to n = 2 (emitting a photon of energy 1.89 eV) and then a transition from n = 2 to n = 1 (emitting another photon of energy 10.2 eV). These two possibilities are shown in Figure 21.5.

Figure 21.5: Transitions from n = 3 in hydrogen.

Whether the electron will choose to make the direct or the indirect transition is an issue of chance: there is a probability for the one option and another probability for the other. (The Bohr model cannot predict these probabilities, but the theories that followed years later can.)

So the emission spectra of elements can be understood if we accept that electrons in atoms exist in energy levels with discrete energy.

CHECK YOURSELF 3

When does an atom emit a photon?

Absorption spectra

Now imagine sending a beam of white light through a gas. The majority of the atoms in the gas are in their ground state. Electrons in the atoms may absorb photons in the beam and move to an excited state. This will happen only if the photon that is to be absorbed has exactly the right energy that corresponds to the difference in energy between the ground state and an excited state. This means that the light that is transmitted through the gas will be missing the photons that have been absorbed. This gives rise to absorption spectra the set of wavelengths that can be absorbed (Figure 21.6). The dark lines correspond to the wavelengths of the absorbed photons. They are at the same wavelengths as the emission spectra.

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Figure 21.6: The absorption spectrum of hydrogen (top) and the emission spectrum (bottom). The emission lines and the absorption lines are at the same wavelength.

CHECK YOURSELF 4

Figure 21.7 shows four atomic transitions.

Figure 21.7: For check yourself question 4.

Which one results in the emission of a photon with the largest wavelength?

Exam tip

The electrons that absorb photons will move to an excited state, but once there, they will make a down transition, emitting the photons they absorbed. So why are the photons missing? This is because the photons are emitted in all directions and not necessarily along the direction the observer is looking. You must be able to explain this in an exam.

Experiments show that no two elements have the same wavelengths in their emission and absorption spectra. These wavelengths are like fingerprints—they are unique to each element and so can be used to identify the element. This has been used to identify the chemical composition of stars.

Worked Example 21.2

Calculate the wavelength of the photon emitted in the transition from the first excited level to the ground state of mercury.

Answer

From Figure 21.4b the energy difference is:

−5.77−(−10.44)=4.67 eV

So the wavelength is found from:

h c λ=4.67 eV=4.67×1.6×10−19 J

λ=6.63×10−34×3.0×10 8 7.472×10−19

λ=2.7×10−7 m

This is an ultraviolet wavelength and so does not show up in the emission spectrum in Figure 21.1.

Atoms are very particular when it comes to absorbing energy! If the energy is provided by a photon, the energy will be accepted only if the photon energy corresponds to exactly a difference of energies between levels. If the energy is provided by an incoming electron, then the atom may accept what is needed to make a transition to a higher level and the electron then bounces off with an energy reduced by the energy absorbed.

CHECK YOURSELF 5

An electron is in the ground state of hydrogen. The difference in energy between the ground state and the first excited state is 10.2 eV. What will happen to this electron if a a photon and b an electron each of energy 10.3 eV is incident on the atom?

Links

E. Hubble discovered that spectral lines emitted from distant galaxies show a redshift, i.e. the measured wavelength on earth is longer than that emitted. This led to the notion of the expanding universe and Hubble’s law. Measuring the redshift allows a measurement of speed and through Hubble’s law the distance to the galaxy. Also, as mentioned earlier, study of the spectral lines of stars allows the determination of the elements in the star.

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NATURE OF SCIENCE

Figure 21.8: Ernest Rutherford.

Ernest Rutherford (Figure 21.8) was a very practical man with a cheerful personality. For his achievements he was compared to Newton—they are buried near each other in Westminster Abbey, London.

Rutherford has many famous quotes, such as:

‘Don’t let me catch anyone talking about the Universe in my department’.

‘All of Physics is either impossible or trivial. It is impossible until you understand it, then it becomes trivial’.

‘Equipment costs money, so we have to use our heads’.

‘A theory you can’t explain to a bartender is probably no damn good’.

Of the many stories about Rutherford, one involves the Russian physicist Pyotr Kapitza who wanted to become Rutherford’s student. Rutherford declined, saying he had more than 10 students already and had no room for another. Kapitza asked Rutherford what error he was willing to accept in his experiments, to which Rutherford replied 10%. ‘Then you can take me’ said Kapitza. Rutherford did! Five of Rutherford’s students, including Kapitza, went on to receive Nobel prizes in physics.

TEST YOUR UNDERSTANDING

1What is the approximate value of the ratio of the radius of an atom to the radius of its nucleus?

2Name the force that keeps electrons attached to the nucleus in Rutherford’s model of the atom.

3In the gold foil experiment explain why:

athe foil was very thin

bthe experiment was done in an evacuated container.

4 aDiscuss what is meant by the statement that the energy of atoms is discrete.

bOutline the evidence for this discreteness.

5Explain why the dark lines of an absorption spectrum have the same wavelengths as the bright lines of an emission spectrum for the same element.

6What is the energy of a photon whose wavelength is 480 nm?

7The energy of a photon is 3.8 eV. What is its wavelength?

8Calculate the wavelength of the photon emitted in a transition from n = 4 to n = 2 in hydrogen.

9Refer to Figure 21.3. Explain why the distance between the emission lines of hydrogen decreases as we move to the right.

10A hydrogen atom is in its ground state.

aExplain the term ground state.

bPhotons of energy 10.4 eV are directed at hydrogen gas in its ground state. Suggest what, if anything, will happen to the hydrogen atoms.

cIn another experiment, a beam of electrons of energy 10.4 eV are directed at hydrogen atoms in their ground state. Suggest what, if anything, will happen to the hydrogen atoms and the electrons in the beam.

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11 aWhat is the evidence for the existence of energy levels in atoms?

bElectrons of kinetic energy:

i10.10 eV

ii12.80 eV

iii13.25 eV

collide with hydrogen atoms and can excite these to higher states.

In each case, find the largest n corresponding to the state the atom can be excited to. Assume that the hydrogen atoms are in their ground state initially.

12 aWhat do you understand by the term ionisation energy?

bWhat is the ionisation energy for a hydrogen atom in the state n = 3?

13What is the minimum speed an electron must have so that it can ionise an atom of hydrogen in its ground state?