Nuclear Radius

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Rutherford Scattering Experiment

Rutherford first determined the radius of the nucleus in his famous alpha particle scattering experiment. The answer he arrived at was quite a surprise.

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Change in energy

  • The alpha particles in Rutherford’s experiment had a kinetic energy of 5.3 MeV.
  • The gold nucleus in the experiment does work on the alpha particle to slow it down. Eventually, the alpha particle has zero kinetic energy (KE) and then is repelled.
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Distance of closest approach

  • The closest point of approach is where the KE lost = electrostatic potential energy gained.
  • The formula for electrostatic potential energy is:
    • PE =Qq4πϵ0r = \frac{Qq}{4\pi \epsilon_0r} where r is the distance of separation of charges Q and q.
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Distance of closest approach 2

  • So:
    • (5.3×106)×(1.6×1019)=(79×1.6×1019)(2×1.6×1019)4πϵ0r(5.3 \times 10^6) \times (1.6 \times 10^{-19}) = \frac{(79 \times 1.6 \times 10^{-19})(2 \times 1.6 \times 10^{-19})}{4\pi \epsilon_0 r}
  • Which gives:
    • r=79×2×(1.6×1019)(5.3×106)×(4πϵ0)=4.3×1014 r = \frac{79 \times 2 \times (1.6 \times 10^{-19})}{(5.3 \times 10^6) \times (4\pi \epsilon_0)} = 4.3 \times 10^{-14} m
  • This is an over-estimate for the radius of the nucleus. Rutherford was convinced that the alpha particles had not come into contact with the gold nucleus.

Electron Diffraction Experiment

Electron diffraction experiments are an alternative method of finding the radius of the nucleus.

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De Broglie wavelength

  • The electron can behave like a wave, with de Broglie wavelength given by:
    • λ=hp\lambda =\frac{h}{p}
      • Where h is the Planck constant and p is the momentum of the electron.
  • The de Broglie wavelength is a wavelength associated with matter based on its mass and momentum.
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Intensity of electrons

  • The graph of intensity of diffracted electrons against diffraction angle is shown here.
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Explanation of graph

  • The angle θ, where the intensity is a minimum, can be explained by assuming that the electrons have diffracted around a spherical object (i.e. the nucleus).
  • The diameter of the nucleus, d, can be found using the equation:
    • sinθ=1.22λd\sin \theta = 1.22 \frac{\lambda}{d}

Liquid Drop Model of the Nucleus

We assume the nucleus is of constant density and spherical in the liquid drop model.

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  • The liquid drop model of the nucleus means that the volume of the nucleus is directly proportional to the number of nucleons present in that nucleus and that the nucleus is spherical.
  • V=kA=43πr3V = kA = \frac43 \pi r^3
  • Where r is the radius of the nucleus with mass number A and volume V.
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  • Using the previous equation:
    • r=3kA4π3r = \sqrt[3]{\frac{3kA}{4\pi}}
  • Or more conveniently:
    • r=r0A1/3r=r_0A^{1/3}
    • Where r0 is the radius of the proton.
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Validity of model

  • The idea on the previous slide and the subsequent experimental confirmation is evidence that the liquid drop model is correct.
  • The consequence of this is that nuclear material has a constant density.

Jump to other topics

1Measurements & Errors

2Particles & Radiation


4Mechanics & Materials


6Further Mechanics & Thermal Physics (A2 only)

7Fields & Their Consequences (A2 only)

8Nuclear Physics (A2 only)

9Option: Astrophysics (A2 only)

10Option: Medical Physics (A2 only)

11Option: Engineering Physics (A2 only)

12Option: Turning Points in Physics (A2 only)

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