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Although the results of the experiment were published by his colleagues in 1909, it took Rutherford two years to convince himself of their meaning. Like Thomson before him, Rutherford was reluctant to accept such radical results. Nature on a small scale is so unlike our classical world that even those at the forefront of discovery are sometimes surprised. 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. The size of the nucleus was determined to be about 10 15 m size 12{"10" rSup { size 8{ - "15"} } " m"} {} , or 100,000 times smaller than the atom. This implies a huge density, on the order of 10 15 g/cm 3 size 12{"10" rSup { size 8{"15"} } " g/cm" rSup { size 8{3} } } {} , vastly unlike any macroscopic matter. Also implied is the existence of previously unknown nuclear forces to counteract the huge repulsive Coulomb forces among the positive charges in the nucleus. Huge forces would also be consistent with the large energies emitted in nuclear radiation.

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

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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Questions & Answers

what is a half life
Mama Reply
the time taken for a radioactive element to decay by half of its original mass
ken
what is radioactive element
mohammed
Half of the total time required by a radioactive nuclear atom to totally disintegrate
Justice
radioactive elements are those with unstable nuclei(ie have protons more than neutrons, or neutrons more than protons
Justice
in other words, the radioactive atom or elements have unequal number of protons to neutrons.
Justice
state the laws of refraction
Fabian
state laws of reflection
Fabian
Why does a bicycle rider bends towards the corner when is turning?
Mac
When do we say that the stone thrown vertically up wards accelerate negatively?
Mac
Give two importance of insulator placed between plates of a capacitor.
Mac
Macho had a shoe with a big sole moving in mudy Road, shanitah had a shoe with a small sole. Give reasons for those two cases.
Mac
when was the name taken from
Biola Reply
retardation of a car
Biola
when was the name retardation taken
Biola
did you mean a motion with velocity decreases uniformly by the time? then, the vector acceleration is opposite direction with vector velocity
Sphere
Atomic transmutation
Basirat Reply
An atom is the smallest indivisible particular of an element
mosco Reply
what is an atomic
Awene Reply
reference on periodic table
Titus Reply
what Is resonance?
Mozam Reply
phenomena of increasing amplitude from normal position of a substance due to some external source.
akif
What is a black body
Amey Reply
Black body is the ideal body can absorb and emit all radiation
Ahmed
the emissivity of black body is 1. it is a perfect absorber and emitter of heat.
Busayo
Why is null measurement accurate than standard voltmeter
Neemat Reply
that is photoelectric effect ?
Sabir Reply
It is the emission of electrons when light hits a material
Anita
Yeah
yusuf
is not just a material
Neemat
it is the surface of a metal
Neemat
what is the formula for time of flight ,maxjmum height and range
agangan Reply
what is an atom
Awene
how does a lightning rod protect a building from damage due to lightning ?
Faith Reply
due to its surface lustre but due to some factors it can corrode but not easily as it lightning surface
babels
pls what is mirage
babels
light rays bend to produce a displaced image of distant objects; it's an natural & optical phenomenon......
Deepika
what is the dimensional formula for torque
Otto Reply
L2MT-2
Jolly
same units of energy
Baber
what is same units of energy?
Baber
Nm
Sphere
Ws
Sphere
CV
Sphere
M L2 T -2
Dokku
it is like checking the dimension of force. which is ML2T-2
Busayo
ML2T-2
Joshua
M L2 T-2
Samuel
what is the significance of moment of inertia?
study
an object of mass 200g moves along a circular path of radius 0.5cm with a speed of 2m/s. calculate the angular velocity ii period iii frequency of the object
Faith Reply
w = 2/(0.005) period = PIE(0.005) f = 1/(PIE(0.005)) assuming uniform motion idk..
Georgie
w=2/(0.005)×100
isaac
supposed the speed on the path is constant angular velocity w (rad/s) = v (m/s) : R (m) period T (s) = 2*Pi * R : v frequency f ( Hz) = 1: T
Sphere
a=w.w.r=mv.v/r,w=mv/r=0.2×2/0.005=80rads-s
Mac
in the pole vaulter problem, how do they established that the mass is 5.00kg? where did that number come from?
-- Reply
Practice Key Terms 2

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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