<< Chapter < Page Chapter >> Page >

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.

Got questions? Get instant answers now!

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.

Got questions? Get instant answers now!

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} } } {}

Got questions? Get instant answers now!

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.

Got questions? Get instant answers now!

(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"} {} .

Got questions? Get instant answers now!

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
Aislinn Reply
cm
tijani
what is titration
John Reply
what is physics
Siyaka Reply
A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
Jude Reply
Can you compute that for me. Ty
Jude
what is the dimension formula of energy?
David Reply
what is viscosity?
David
what is inorganic
emma Reply
what is chemistry
Youesf Reply
what is inorganic
emma
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
Adjei
please, I'm a physics student and I need help in physics
Adjanou
chemistry could also be understood like the sexual attraction/repulsion of the male and female elements. the reaction varies depending on the energy differences of each given gender. + masculine -female.
Pedro
A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
Krampah Reply
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
Sahid Reply
you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
Ryan
what's motion
Maurice Reply
what are the types of wave
Maurice
answer
Magreth
progressive wave
Magreth
hello friend how are you
Muhammad Reply
fine, how about you?
Mohammed
hi
Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
yasuo Reply
Who can show me the full solution in this problem?
Reofrir Reply
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply
Practice Key Terms 2

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'College physics' conversation and receive update notifications?

Ask