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The small size of the nucleus also implies that the atom is mostly empty inside. In fact, in Rutherford’s experiment, most alphas went straight through the gold foil with very little scattering, since electrons have such small masses and since the atom was mostly empty with nothing for the alpha to hit. There were already hints of this at the time Rutherford performed his experiments, since energetic electrons had been observed to penetrate thin foils more easily than expected. [link] shows a schematic of the atoms in a thin foil with circles representing the size of the atoms (about 10 10 m size 12{"10" rSup { size 8{ - "10"} } " m"} {} ) and dots representing the nuclei. (The dots are not to scale—if they were, you would need a microscope to see them.) Most alpha particles miss the small nuclei and are only slightly scattered by electrons. Occasionally, (about once in 8000 times in Rutherford’s experiment), an alpha hits a nucleus head-on and is scattered straight backward.

The image shows an enlarged view of atoms in gold foil having a diameter of ten to the power minus ten meter and a dot within it representing the nucleus. A few alpha rays are shown passing through the atoms. Some are scattered as they hit the nuclei while some are just passing through.
An expanded view of the atoms in the gold foil in Rutherford’s experiment. Circles represent the atoms (about 10 10 m size 12{"10" rSup { size 8{ - "10"} } " m"} {} in diameter), while the dots represent the nuclei (about 10 15 m size 12{"10" rSup { size 8{ - "15"} } " m"} {} in diameter). To be visible, the dots are much larger than scale. Most alpha particles crash through but are relatively unaffected because of their high energy and the electron’s small mass. Some, however, head straight toward a nucleus and are scattered straight back. A detailed analysis gives the size and mass of the nucleus.

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, with mostly vacuum inside the atom. This picture is analogous to how low-mass planets in our solar system orbit the large-mass Sun at distances large compared with the size of the sun. In the atom, the attractive Coulomb force is analogous to gravitation in the planetary system. (See [link] .) Note that a model or mental picture is needed to explain experimental results, since the atom is too small to be directly observed with visible light.

The image shows three elliptical orbits showing electrons’ movement around a positive nucleus. The movement of the electrons in the orbit shown with arrows are opposite to each other.
Rutherford’s planetary model of the atom incorporates the characteristics of the nucleus, electrons, and the size of the atom. This 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.

Rutherford’s planetary model of the atom was crucial to understanding the characteristics of atoms, and their interactions and energies, as we shall see in the next few sections. Also, it was an indication of how different nature is from the familiar classical world on the small, quantum mechanical scale. The discovery of a substructure to all matter in the form of atoms and molecules was now being taken a step further to reveal a substructure of atoms that was simpler than the 92 elements then known. We have continued to search for deeper substructures, such as those inside the nucleus, with some success. In later chapters, we will follow this quest in the discussion of quarks and other elementary particles, and we will look at the direction the search seems now to be heading.

Phet explorations: rutherford scattering

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

Rutherford Scattering

Test prep for ap courses

In an experiment, three microscopic latex spheres are sprayed into a chamber and become charged with +3 e , +5 e , and −3 e , respectively. Later, all three spheres collide simultaneously and then separate. Which of the following are possible values for the final charges on the spheres? Select two answers.

  1. +4 e , −4 e , +5 e
  2. −4 e , +4.5 e , +4.5 e
  3. +5 e , −8 e , +7 e
  4. +6 e , +6 e , −7 e

(a), (d)

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In Millikan’s oil drop experiment, he experimented with various voltage differences between two plates to determine what voltage was necessary to hold a drop motionless. He deduced that the charge on the oil drop could be found by setting the gravitational force on the drop (pointing downward) equal to the electric force (pointing upward):

m drop g = q E ,

where m drop is the mass of the oil drop, g is the gravitational acceleration (9.8 m/s 2 ), q is the net charge of the oil drop, and E is the electric field between the plates. Millikan deduced that the charge on an electron, e , is 1.6 × 10 −19 C.

For a system of oil drops of equal mass (1.0 × 10 −15 kilograms), describe what value or values of the electric field would hold the drops motionless.

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Section summary

  • Atoms are composed of negatively charged electrons, first proved to exist in cathode-ray-tube experiments, and a positively charged nucleus.
  • All electrons are identical and have a charge-to-mass ratio of
    q e m e = 1.76 × 10 11 C/kg. size 12{ { {q rSub { size 8{e} } } over {m rSub { size 8{e} } } } = - 1 "." "76" times "10" rSup { size 8{"11"} } " C/kg" "." } {}
  • The positive charge in the nuclei is carried by particles called protons, which have a charge-to-mass ratio of
    q p m p = 9 . 57 × 10 7 C/kg . size 12{ { {q rSub { size 8{p} } } over {m rSub { size 8{p} } } } =9 "." "57" times "10" rSup { size 8{7} } " C/kg" "." } {}
  • Mass of electron,
    m e = 9 . 11 × 10 31 kg . size 12{m rSub { size 8{e} } =9 "." "11" times "10" rSup { size 8{ - "31"} } " kg" "." } {}
  • Mass of proton,
    m p = 1 . 67 × 10 27 kg. size 12{m rSub { size 8{p} } =1 "." "67" times "10" rSup { size 8{ - "27"} } " kg"} {}
  • The planetary model of the atom pictures electrons orbiting the nucleus in the same way that planets orbit the sun.

Conceptual questions

What two pieces of evidence allowed the first calculation of m e size 12{m"" lSub { size 8{e} } } {} , the mass of the electron?

(a) The ratios q e / m e size 12{q rSub { size 8{e} } /m rSub { size 8{e} } } {} and q p / m p size 12{q rSub { size 8{p} } /m rSub { size 8{p} } } {} .

(b) The values of q e size 12{q rSub { size 8{e} } } {} and E B size 12{E rSub { size 8{B} } } {} .

(c) The ratio q e / m e size 12{q rSub { size 8{e} } /m rSub { size 8{e} } } {} and q e size 12{q rSub { size 8{e} } } {} .

Justify your response.

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How do the allowed orbits for electrons in atoms differ from the allowed orbits for planets around the sun? Explain how the correspondence principle applies here.

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Problem exercises

Rutherford found the size of the nucleus to be about 10 15 m size 12{"10" rSup { size 8{ - "15"} } " m"} {} . This implied a huge density. What would this density be for gold?

6 × 10 20 kg/m 3 size 12{1 "." "93" times "10" rSup { size 8{"25"} } `"kg/m" rSup { size 8{3} } } {}

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In Millikan’s oil-drop experiment, one looks at a small oil drop held motionless between two plates. Take the voltage between the plates to be 2033 V, and the plate separation to be 2.00 cm. The oil drop (of density 0 . 81 g/cm 3 size 12{0 "." "81 g/cm" rSup { size 8{3} } } {} ) has a diameter of 4 . 0 × 10 6 m size 12{4 "." 0 times "10" rSup { size 8{ - 6} } " m"} {} . Find the charge on the drop, in terms of electron units.

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(a) An aspiring physicist wants to build a scale model of a hydrogen atom for her science fair project. If the atom is 1.00 m in diameter, how big should she try to make the nucleus?

(b) How easy will this be to do?

(a) 10.0 μm size 12{"10" "." 0" μm"} {}

(b) It isn’t hard to make one of approximately this size. It would be harder to make it exactly 10.0 μm size 12{"10" "." 0" μm"} {} .

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Practice Key Terms 2

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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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