Nuclear Physics 6 Nuclear Radius Saturday, 08 December 2018 Leeds City College

Rutherford Scattering Limitations: The nucleus is treated as a point charge. At this level it is not. The alpha particles are stopped some distance away from the nucleus. It takes higher energy alpha particles to penetrate the nucleus. The values for the nuclear radius given by other particles such as protons, neutrons and electrons are slightly different. Gold nucleus rC Alpha particle stationary at P Alpha coming in P Saturday, 08 December 2018 Leeds City College

Estimating the size of the Nucleus Kinetic energy of the alpha particle = eV Charge of the alpha particle = 2 × 1.6 × 10-19 C Energy required to get to distance r from the nucleus: Charge of the gold nucleus = 79 × 1.6 × 10-19 C Energy of the alpha particle = 5 MeV = 5 × 106 × 1.6 × 10-19 J What is the closest approach? Convert the energy of the alpha particle from eV to J by multiplying by 1.6 × 10-19 . r = 9 × 109 × 2 × 1.6 × 10-19 × 79 × 1.6 × 10-19 = 4.6 × 10-14 m 8 × 10-13 Saturday, 08 December 2018 Leeds City College

Electron Diffraction The electrons are accelerated to a very high speed to give a de Broglie wavelength of about 10-15 m (1 fm) Saturday, 08 December 2018 Leeds City College

Calculating the de Broglie wavelength From unit 1 we saw that the de Broglie wavelength was given by: We could combine the expressions by squaring the de Broglie equation, but it would be messy. So keep them apart and work out v, before substituting into the de Broglie equation. We can work out the speed by: What is the speed of electrons to give a wavelength of 1 × 10-15 m? Saturday, 08 December 2018 Leeds City College

Relativistic Effects Rearrange the de Broglie equation: v = 6.63 × 10-34 = 7.28 × 1011 m/s 9.11 × 10-31 × 1 × 10-15 Since nothing can travel faster than the speed of light (3 × 108 m/s), the method is clearly wrong. As electrons approach the speed of light, they gain mass due to the effects of relativity. This is beyond what is required on the syllabus. Saturday, 08 December 2018 Leeds City College

Angle of Diffraction The relationship that is observed is: This can be compared to the relationship for the diffraction of light: The factor of 0.61 arises from the diffraction around a spherical object. Saturday, 08 December 2018 Leeds City College

Nuclear Radius We need to remember that the nucleus has a boundary that is rather fuzzy. However at this level, we can assume that the nucleus is spherical. Equation: A1/3 = 3A The data that give this equation have been obtained for many different nuclides. The term r0 is a constant = 1.05 × 10-15 m These graphs show a straight line relationship… Saturday, 08 December 2018 Leeds City College

Graphs ln R R R3 ln r0 ln A A1/3 A Gradient = r0 Gradient = r03 Saturday, 08 December 2018 Leeds City College

Nuclear Density Density = 3.43 × 1017 kg m-3 Assume that the nucleus is spherical with radius A. It has a mass of 1 atomic mass unit. It consists of 1 nucleon. To work out the radius we use: 1 u = 1.661 × 10-27 kg Radius = 1.05 × 10-15 m Density = mass = 1.661 × 10-27 volume 4.85 × 10-45 Volume: Volume = 4/3 × p × (1.05 × 10-15 m)3 = 4.85 × 10-45 m3 Density = 3.43 × 1017 kg m-3 1 mm3 will have a mass of 340 000 tonnes. Drop it on your foot and it will bring tears to your eyes. Saturday, 08 December 2018 Leeds City College

Density of a neutron star This neutron star has a mass about twice that of the Sun. Mass = 4 × 1030 kg. Its diameter is about 25 km. Its density is 6 × 1016 kg m-3, which is less than the density of nuclear material. Fraser Cain Saturday, 08 December 2018 Leeds City College