1. nuclear binding energy emre özyetiş 10-U

2. The binding energy is the energy required to decompose the nucleus into protons and neutrons.

3. In chemical reactions, the nucleus remains the same; no new atoms are formed. The number and kinds of atoms are conserved. There is no extra energy given off from nucleus besides the energy given off or taken in by the breakage or forming of bonds.

4. However, in nuclear reactions, divisions and combinations of nuclei occur. New atoms are formed; rays and particles with energy are emitted.

5. When new atoms are formed, binding energy is given off to keep the neutrons and protons together and to become stable. This energy can be calculated by Einstein’s famous formula. E=mc 2 energy=mass×(speed of light) 2

6. According to The Relativity Theory of Einstein, some of the mass is converted into energy and emitted so that more stable atoms are formed. This is why in nuclear reactions energy and mass are conserved together.

7. ΔE=Δmc 2 Where Δ E is binding energy Δm is mass defect: (m n 0 + m p + )- m element itself c is speed of light in vacuum which is 2.99×10 8 m×s -1 When units are kg for mass defect and m.s -1 for speed of light, then we get kg×m 2 ×s -2 which is Joule.

8. To make comparison between different nuclides easily, we express binding energies on a per-nucleon. Example.1b will give information about BE/nucleon

9. Mass of n= 1.67493×10 -24 g Mass of p= 1.67262×10 -24 g Mass of O nucleus is 2.65535×10 -23 g Example.1: Calculate a) the change in energy if 1 mol O nuclei were formed from neutrons and protons. b) Binding Energy per nucleon. 16 8  For Δm we ought to know the equation of formation 8 n + 8 p O 1010 1111 16 8 1010 1111 16 8

10. In nucleus the difference is= (Mass of O) – (Mass of 8p and 8n) = -2.269×10 -25 g The difference in 1 mole is = (6.022×10 23 nuclei/mol) × -2.269×10 -25 g/nucleus Δm = -0.1366 g/mol ΔE= Δmc 2 ΔE= -0.1366 g/mol × (3.00×10 8 m/s) 2 × 1 kg/1000g ΔE= -1.23×10 13 J/mol a)

11. b) In order to find ΔE per nucleon, we need to know ΔE per nucleus first ΔE per O nucleus = =-2.04×10 -11 J/nucleus -1.23×10 13 J/mol 6.022×10 23 nuclei/mol In terms of a more convenient energy unit, a million electron volts where; 1 MeV = 1.60 × 10 -13 J -2.04×10 -11 J/nucleus =-1.28 × 10 2 MeV/nucleus

12. At last we can calculate ΔE per nucleon by dividing by the sum of neutrons and protons; which is 16 for O ΔE per nucleon for O is 16 8 -1.28 × 10 2 MeV/nucleus 16 nucleons/nucleus = =-7.98 MeV/nucleon 16 8

13. Example.2: In the radioactive decomposition of radium-226, the equation for the nuclear process is Ra Rn + He How much mass is converted into energy during this radioactive decay process? Ra: 226.025360 amu Rn: 222.017530 amu He: 4.00260361 amu 226 88 222 86 4242

14. Δm= m produced -m reacted Δm =( 222.017530amu + 4.00260361amu ) - 226.025360amu Δm= -5.226×10 -3 amu is lost in this process. This means that amount of mass is converted into energy ΔE=Δmc 2 ΔE= (-5.226×10-3 g/mol) × 1 kg/1000g × (3.00×10 8 m/s) 2 ΔE= -4.70×10 11 J/mol Negative sign indicates that the process is exothermic

16. The binding energy curve is obtained by dividing the total nuclear binding energy by the number of nucleons. The fact that there is a peak in the binding energy curve in the region of stability near iron means that either the breakup of heavier nuclei (fission) or the combining of lighter nuclei (fusion) will yield nuclei which are more tightly bound (less mass per nucleon).