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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 happens when an unstoppable force collides an immovable object?
Mavis Reply
a radioactive nuclei of mass 6.0g has a half life of 8days. calculate during which 5.25g of the nuclei would have decay
ADEMOLA Reply
Calculate the Newton's the weight of a 2.5 Kilogram of melon. What is its weight in pound?
Rialyn Reply
calculate the tension of the cable when a buoy with 0.5m and mass of 20kg
Iga Reply
what is displacement
Nyamza Reply
it's the time rate of change of distance
Mollamin
distance in a given direction is diplacement
Musa
Distance in a spacified direction
Gift
you shouldn't say distance,displacement and distance are two different things .distance can be lopped curved but displacement is always in a straight line so you can't use distance to define it. displacement is the change of position in a specified direction.
Joshua
Well stayed josh👍
Gift
thank you gift.
Joshua
well explained
Mary
what is the meaning of physics
Alausa Reply
to study objects in motion and how they interact or take part in the natural phenomenon of the universe.
Phill
an object that has a small mass and an object has a large mase have the same momentum which has high kinetic energy
Faith Reply
The with smaller mass
Gift
how
Faith
Since you said they have the same momentum.. So meaning that there is more like an inverse proportionality in the quantities used to find the momentum. We are told that the the is a larger mass and a smaller mass., so we can conclude that the smaller mass had higher velocity as compared to other one
Gift
Mathamaticaly correct
megavado
Mathmaticaly correct :)
megavado
I have proven it by using my own values
Gift
Larger mass=4g Smaller mass=2g Momentum of both=8 Meaning V for L =2 and V for S=4 Now find there kinetic energies using the data presented
Gift
grateful soul...thanks alot
Faith
Welcome
Gift
2 stones are thrown vertically upward from the ground, one with 3 times the initial speed of the other. If the faster stone takes 10 s to return to the ground, how long will it take the slower stone to return? If the slower stone reaches a maximum height of H, how high will the faster stone go
Julliene Reply
30s
Gift
how can i calculate it's height
Julliene
is speed the same as velocity
Faith Reply
no
Nebil
in a question i ought to find the momentum but was given just mass and speed
Faith
just multiply mass and speed then you have the magnitude of momentem
Nebil
Yes
Gift
Consider speed to be velocity
Gift
it worked our . . thanks
Faith
Distinguish between semi conductor and extrinsic conductors
Okame Reply
Suppose that a grandfather clock is running slowly; that is, the time it takes to complete each cycle is longer than it should be. Should you (@) shorten or (b) lengthen the pendulam to make the clock keep attain the preferred time?
Aj Reply
I think you shorten am not sure
Uche
shorten it, since that is practice able using the simple pendulum as experiment
Silvia
it'll always give the results needed no need to adjust the length, it is always measured by the starting time and ending time by the clock
Paul
it's not in relation to other clocks
Paul
wat is d formular for newton's third principle
Silvia
okay
Silvia
shorten the pendulum string because the difference in length affects the time of oscillation.if short , the time taken will be adjusted.but if long ,the time taken will be twice the previous cycle.
FADILAT
discuss under damped
Prince Reply
resistance of thermometer in relation to temperature
Ifeanyi Reply
how
Bernard
that resistance is not measured yet, it may be probably in the next generation of scientists
Paul
Is fundamental quantities under physical quantities?
Igwe Reply
please I didn't not understand the concept of the physical therapy
John Reply
physiotherapy - it's a practice of exercising for healthy living.
Paul
what chapter is this?
Anderson
this is not in this book, it's from other experiences.
Paul
am new in the group
Daniel
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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