logearav Posted November 25, 2011 Posted November 25, 2011 Revered Members, Binding energy for lighter nuclei is large and for heavier nuclei BE is less. Why? I know BE is the energy needed to bind protons and neutrons in a nucleus. Also it can be defined as the energy that must be supplied to disrupt the nucleus into constituent protons and neutrons. Take a lighter nuclei which has 3 protons and so ideally the energy needed to bind these 3 protons should be less when compared with the energy needed to bind 82 protons(say) in a heavier nuclei . So, why BE is less for heavier nuclei than lighter nuclei?
Klaynos Posted November 25, 2011 Posted November 25, 2011 I'd suggest the best way to start understanding this is reading: http://en.wikipedia.org/wiki/Liquid-drop_model 1
swansont Posted November 25, 2011 Posted November 25, 2011 Revered Members, Binding energy for lighter nuclei is large and for heavier nuclei BE is less. Why? I know BE is the energy needed to bind protons and neutrons in a nucleus. Also it can be defined as the energy that must be supplied to disrupt the nucleus into constituent protons and neutrons. Take a lighter nuclei which has 3 protons and so ideally the energy needed to bind these 3 protons should be less when compared with the energy needed to bind 82 protons(say) in a heavier nuclei . So, why BE is less for heavier nuclei than lighter nuclei? I's not. Total binding energy increases as nuclei get bigger. The average binding energy per nucleon, however, peaks near A=60, meaning that splitting large nuclei or fusing small ones tends to release energy. Also, BE is not the energy needed to bind protons and neutrons in a nucleus. It's the energy released when doing so. A bound system has less energy than the free constituent particles. 1
mississippichem Posted November 25, 2011 Posted November 25, 2011 Also, BE is not the energy needed to bind protons and neutrons in a nucleus. It's the energy released when doing so. A bound system has less energy than the free constituent particles. Swansont, A small aside. Is the difference between the sum of the masses of the free nucleons and the mass of the bound nucleus referred to as "mass defect" or "mass deficit"? Basically I'm wondering if the two terms are interchangable.
swansont Posted November 25, 2011 Posted November 25, 2011 Swansont, A small aside. Is the difference between the sum of the masses of the free nucleons and the mass of the bound nucleus referred to as "mass defect" or "mass deficit"? Basically I'm wondering if the two terms are interchangable. I was taught mass defect, but I've seen deficit on occasion. AFAIK they are interchangeable — I am not aware of one term have some other meaning.
logearav Posted November 30, 2011 Author Posted November 30, 2011 Also, BE is not the energy needed to bind protons and neutrons in a nucleus. It's the energy released when doing so. A bound system has less energy than the free constituent particles. Thanks for the reply Swansont. Wikipedia says "At the nuclear level, binding energy is also equal to the energy liberated when a nucleus is created from other nucleons or nuclei. ". Why the energy released is more for binding nucleons for smaller nuclei when compared to heavier nuclei"
swansont Posted November 30, 2011 Posted November 30, 2011 Thanks for the reply Swansont. Wikipedia says "At the nuclear level, binding energy is also equal to the energy liberated when a nucleus is created from other nucleons or nuclei. ". Why the energy released is more for binding nucleons for smaller nuclei when compared to heavier nuclei" It's not. Nuclei generally have a greater binding energy as they get larger numbers of nucleons. The binding energy for He-4, for example, is about 28 MeV. The binding energy for Uranium is more than 1.5 GeV (i.e. > 1500 MeV)
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