4%20Nuclear%20Fission.ppt

First term is: There then follows a series of “correction terms” to take into account the binding energy. Since BE/nucleon is roughly constant then the main correction term will just be proportional to the number of nucleons A. The second term is then the volume energy -avA. The nucleons at the surface will, however, only be attracted by nucleons on one side so that the BE term in 3 will be an exaggeration. This must be corrected by a term proportional to the surface area +asA2/3 surface energy(of that of liquid droplet) The Z protons in the nucleus will be mutually repelled resulting in a negative binding. Assuming Z protons uniformly distributed throughout nuclear volume a simple classical calculation would give a potential energy for the system of We put this in the formula as Coulomb Energy Experimental evidence shows that BE/nucleon is a maximum for light nuclei when Z = N. This means that if Z ≠ A/2 the nucleus is less strongly bound than it might be. This lack of binding energy will be proportional to the number of pairs of nucleons in excess: In fact a reasonable form for this term is: A symmetry energy Most stable nuclei are even-even Even A even Z even N Then…. For Even A odd Z odd N there will be nucleons which cannot pair with another with opposite spin. Such a state is less bound than the even-even case. Thus we need to add a pairing term which is: δ(A,Z) = +ap A-3/4 A even, odd-odd δ(A,Z) = 0 A odd δ(A,Z) = -ap A-3/4 A even, even-even The form of δ(A,Z) not given by liquid drop model. δ(A,Z) has a value ~0.5 pairing energy Total formula: The constants may be obtained empirically by fitting the equation to known masses. Relative Importance of the contributions to Nuclear Binding Energy A MeV/nuc Volume energy Surface energy Coulomb energy Symmetry energy Nuclear Fission Nuclear Fission Binding Energy Survey of Binding Energies Liquid Drop Model Nuclear Mass One mass unit = 1 u (unified mass unit) = 1/12 mass of