Binding Energy

1. Ca Mass Number
2. Protons Of Calcium
3. Calcium Atomic Mass

20 Ca Calcium 40.078. Atomic Number: 20. Atomic Weight: 40.078. Melting Point: 1115 K (842°C or 1548°F) Boiling Point: 1757 K (1484°C or 2703°F) Density: 1.54 grams per cubic centimeter. Phase at Room Temperature: Solid. Element Classification: Metal. Period Number: 4. Group Number: 2. Group Name: Alkaline Earth Metal. What's in a name? California has been hit with devastating wildfires and other natural disasters in both the northern and southern parts of the state. If you need more information about recovery or resources visit the following resources: response.ca.gov wildfirerecovery.org disasterassistance.gov. The isotopic mass data is from G. 20 Calcium 40 Ca 39.962591 96.941 42 Ca 41.958618 0.647 43 Ca 42.958767 0.135.

Ca Mass Number

Thenuclear source of energy is understood only if the concept of binding energy(BE) is understood. The nucleons, as said, are held by short-range nuclearforces. The mass of a nucleus is generally less than the sum of the masses ofthe nucleons. This difference is called the mass-defect.Such a phenomenon is observed in atomic or subatomic levels, and not withordinary materials that we handle, where the total mass has to be the sum of thecomponent masses. The mass-defect is responsible for the binding together of thenucleons. The energy equivalent of the mass-defectis called the binding-energy of the nucleus. Look at the followingillustrative picture and visualize that, when individual nucleons packthemselves together to form a system, which we call nucleus, they liberate someenergy in the form of photons (gamma rays). The liberated energy manifests as areduction in the net mass of the nucleus, as per Einstein’s relation. It mayalso be said that the nucleus could be split intonucleons by supplying external energy equal to the binding energy.

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Example: For the He-4 nucleus, the binding energy may be computed as below: Total mass of 2 protons= 2 ´ 1.00728 = 2.01456 amu Total mass of 2 neutrons = 2 ´ 1.00866 = 2.01732 amu Total mass of nucleons= 4.03188 amu Mass of Alpha particle (measured)= 4.00153 amu Mass defect Dm = 0.03035 amu Hence, Binding energy of He-4 = 28.3 MeV. ( i.e. 0.03035 amu ´ 931.5MeV/amu)

The enormity of the nuclear binding energy can be betterappreciated by comparing it to the binding energy of an electron in an atom. Thenuclear binding energies are on the order of a million times greater than theelectron binding energies of atoms. For example, the bindingenergy of an electron to the nucleus in the hydrogen atom is 13.6 eV,whereas the binding energy ofnucleons in the He-4 nucleus is28,300,000eV (roughly 2 million times!)

BindingEnergy Per Nucleon (BE/A): As the number of particles in a nucleusincreases, the total binding energy also increases. The rate of increase,however, is not uniform. This lack of uniformity results in a variation in theamount of binding energy associated with each nucleon within the nucleus. Thisvariation in the binding energy per nucleon (BE/A) is easily seen when theaverage BE/A is plotted versus atomic mass number (A), as shown in the figurebelow:

Binding Energy per Nucleon vs. Mass Number

The above figure illustrates that as theatomic mass number increases, the binding energy per nucleon decreases for A> 60. The BE/A curve reaches a maximum value of 8.79 MeV at A = 56 anddecreases to about 7.6 MeV for A = 238.

The general shape of the BE/A curve can beexplained using the general properties of nuclear forces. As already said, veryshort-range attractive forces that exist between nucleons hold the nucleus inits shape. On the other hand, the long-range repulsive electrostatic (coulomb)forces that exist between all the protons in the nucleus continuously tend toforce the protons apart, thus splitting the nucleus.As the atomic number (Z) increases, therepulsive electrostatic forces within the nucleus increase. To overcome thisincreased repulsion and maintain stability, the proportion of neutrons in thenucleus must increase. This increase in theneutron-to-proton ratio only partially compensates for the growing proton-protonrepulsive force in the heavier, naturally occurring elements. As therepulsive forces are increasing, less energy must be supplied, on the average,to remove a nucleon from the nucleus. In other words, the BE/A has decreased.The BE/A of a nucleus is an indication of its degree of stability.Generally, the more stable nuclides havehigher BE/A than the less stable ones. The increase in the BE/A asthe atomic mass number decreases from 260 to 60 is the primary reason for the energyliberation in the fission process. In addition, the increase in the BE/Aas the atomic mass number increases from 1 to 60 is the reason for the energyliberation in the fusion process, which is the opposite reaction offission.

The heaviest nuclei require only a small energy addition for therelatively large coulomb forces to overcome the attractive nuclear forces, inorder to split the nucleus into two halves. Connect to ps4 remotely. Consequently, the heaviestnuclei are easily fissionable compared to lighter nuclei.The addition of the energy required to cause a distortion in the heavynucleus leading to fission, is commonly achieved by adding a neutron into thenucleus. Some nuclei experience chances of undergoing fission, even withoutadding a neutron, such a phenomenon being called ‘Spontaneous fission’.

In the fusion of Deuterium (a hydrogen isotope H-2)and Tritium (another hydrogen isotope H-3), called the D-T fusion, the totalenergy released is 17.6 MeV. In the fission of U-235 by neutron, the energyreleased is about 200 MeV. Though fission energy seems higher, therelease of energy per unit mass of the fuel, is much higher in the case offusion than in fission.More discussion on fission and fusion will be given later.