Lecture33 - Chem 162, Lecture 33, Spring 2011 The Nucleus...

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Chem 162, Lecture 33, Spring 2011 The Nucleus The nucleus contains protons and neutrons, each with a mass of about 1.0 amu = 1.0 u (m(proton) = 1.00728 u, m(neutron) = 1.00866 u). The electrons outside the nucleus have a mass of 0.000548 u ~ 1/1836 u (approximately zero to the nearest integer). Almost all of the mass of each atom is in its nucleus - but the size of the nucleus is very small. Radius of an atom is about 10 -10 m. Radius of nucleus is about 10 -15 - 10 -14 m (R nucleus ~ 1.2(A) 1/3 x10 -15 m. Thus the nucleus must be extremely dense. Assume a fluorine nucleus has a radius of 4 x 10 -15 m. Calculate the density of the fluorine nucleus. Mass = 19 g mol x 1 mol 6.02 x 10 23 nuclei = 3.16 x 10 -23 g Volume = 4 3 π R nucleus 3 = 4 3 π (4 x 10 -15 ) 3 = 2.69 x 10 -43 m 3 = 2.69 x 10 -37 cm 3 Density = mass volume = 3.16 x 10 -23 g 2.69 x 10 -37 cm 3 = 1.2 x 10 14 g cm 3 = 1.2 x 10 14 g cm 3 x 1 kg/1000 g x 1 ton/1000 kg = 1.2 x 10 8 ton cm 3 The nucleus is positively charged, consisting of positive protons and neutral neutrons. One of the basic laws of nature is electrostatic attraction (opposite charges) and repulsion (like charges). 1
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If charges q l and q 2 are separated by distance r, the electrostatic force between them is given by Coulomb’s Law: F = kq l q 2 r 2 In the nucleus, the distance “r” between the protons is very small. So there must be a huge electrostatic repulsive force between the protons. So how can the nucleus exist? The stability of the nucleus is due to the strong force , one of the basic forces of nature. The strong force is an attractive force involving a special interaction of the neutrons and protons. There is no simple equation, analogous to Coulomb’s Law, to express the strong force. However, we can say that it is a very short-range force (operates only at very small distances, as on a nuclear scale), and that the stability of the nucleus depends on the neutron/proton ratio. Consider the following list of stable nuclei for the first 10 elements: 1 1 H 4 2 He 7 3 Li 9 4 Be 10 5 B 12 6 C 14 7 N 16 8 O 19 9 F 20 10 Ne 2 1 H 3 2 He 6 3 Li 11 5 B 13 6 C 15 7 N 17 8 O 21 10 Ne 18 8 O 22 10 Ne Note that with the exception of 1 1 H and 3 2 He, the mass number is at least double the atomic number, meaning that every stable isotope has at least as many neutrons as protons. The neutron proton ratio for the first 20 elements is at least 1:1 for stable 2
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isotopes, meaning there must be at least 1 neutron for each proton to provide the strong force for nuclear stability. As for the exceptions mentioned above,
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Lecture33 - Chem 162, Lecture 33, Spring 2011 The Nucleus...

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