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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

but why is the cyclist relative with the canal .
Emmanuel Reply
It is more difficult to obtain a high-resolution ultrasound image in the abdominal region of someone who is overweight than for someone who has a slight build. Explain why this statement is accurate.
Lanii Reply
What is the frictional forc e between two bodies
Kennedy Reply
it is the force which always opposes the motion of the body
why is the force that oppose motion sir
what is a wave
Williams Reply
wave means. A field of study
what are Atoms
is the movement back and front or up and down
how ?
wave is a disturbance that transfers energy through matter or space with little or no associated mass.
A wave is a motion of particles in disturbed medium that carry energy from one midium to another
an atom is the smallest unit( particle) of an element that bares it's chemical properties
what is electromagnetic induction?
what's boy's law
wave is disturbance that travel through a medium ..maybe in vacuum
How is the de Broglie wavelength of electrons related to the quantization of their orbits in atoms and molecules?
Larissa Reply
How do you convert 0.0045kgcm³ to the si unit?
how many state of matter do we really have like I mean... is there any newly discovered state of matter?
Falana Reply
I only know 5: •Solids •Liquids •Gases •Plasma •Bose-Einstein condensate
Alright Thank you
Which one is the Bose-Einstein
can you explain what plasma and the I her one you mentioned
u can say sun or stars are just the state of plasma
but the are more than seven
list it out I wanna know
what the meaning of continuum
Akhigbe Reply
What state of matter is fire
Thapelo Reply
fire is not in any state of matter...fire is rather a form of energy produced from an oxidising reaction.
Isn`t fire the plasma state of matter?
all this while I taught it was plasma
How can you define time?
Thapelo Reply
Time can be defined as a continuous , dynamic , irreversible , unpredictable quantity .
unpredictable? but I can say after one o'clock its going to be two o'clock predictably!
how can we define vector
I would define it as having a magnitude (size)with a direction. An example I can think of is a car traveling at 50m/s (magnitude) going North (direction)
as for me guys u would say time is quantity that measures how long it takes for a specific condition to happen e.g how long it takes for the day to end or how it takes for the travelling car to cover a km.
what is the relativity of physics
Paul Reply
How do you convert 0.0045kgcm³ to the si unit?
What is the formula for motion
Anthony Reply
V=u+at V²=u²-2as
they are eqns of linear motion
v=u+at s=ut+at^\2 v^=u^+2as where ^=2
Explain dopplers effect
Jennifer Reply
Not yet learnt
Explain motion with types
Acceleration is the change in velocity over time. Given this information, is acceleration a vector or a scalar quantity? Explain.
Alabi Reply
Scalar quantity Because acceleration has only magnitude
acleration is vectr quatity it is found in a spefied direction and it is product of displcemnt
its a scalar quantity
velocity is speed and direction. since velocity is a part of acceleration that makes acceleration a vector quantity. an example of this is centripetal acceleration. when you're moving in a circular patter at a constant speed, you are still accelerating because your direction is constantly changing.
acceleration is a vector quantity. As explained by Josh Thompson, even in circular motion, bodies undergoing circular motion only accelerate because on the constantly changing direction of their constant speed. also retardation and acceleration are differentiated by virtue of their direction in
respect to prevailing force
What is the difference between impulse and momentum?
Momentum is the product of the mass of a body and the change in velocity of its motion. ie P=m(v-u)/t (SI unit is kgm/s). it is literally the impact of collision from a moving body. While Impulse is the product of momentum and time. I = Pt (SI unit is kgm) or it is literally the change in momentum
Or I = m(v-u)
the tendency of a body to maintain it's inertia motion is called momentum( I believe you know what inertia means) so for a body to be in momentum it will be really hard to stop such body or object..... this is where impulse comes in.. the force applied to stop the momentum of such body is impulse..
Calculation of kinetic and potential energy
dion Reply
K.e=mv² P.e=mgh
K is actually 1/2 mv^2
what impulse is given to an a-particle of mass 6.7*10^-27 kg if it is ejected from a stationary nucleus at a speed of 3.2*10^-6ms²? what average force is needed if it is ejected in approximately 10^-8 s?
speed=velocity÷time velocity=speed×time=3.2×10^-6×10^-8=32×10^-14m/s impulse [I]=∆momentum[P]=mass×velocity=6.7×10^-27×32×10^-14=214.4×10^-41kg/ms force=impulse÷time=214.4×10^-41÷10^-8=214.4×10^-33N. dats how I solved it.if wrong pls correct me.
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