Nuclear Binding Energy
[1] Binding Energy
- During The 20th Century, The Popular Albert Einstein Came Up
With The Revolutionary Theory, “The Theory Of Relativity”.
- The Theory Explained That Mass And Energy Are
Interconvertible; The Mass Can Be Converted Into Energy And
Vice-Versa.
- This New Dimension To Physics Helped To Resolve Plenty Of
Unsolved Problems And Formed A Forum For Lot Of New
Theories.
- One Among Them Is The Existence Of Nuclear Binding Energy.
- The Theoretical Explanation For The Mass Defect Is Based On
Einstein’s Equation E = mc^2.
- When The Z Protons And N Neutrons Combine To Make A
Nucleus, Some Of The mass (∆ m) Disappears Because It Is
Converted Into An
Amount Of Energy ∆ E = (∆ m)c^2
- This Energy Is Called The Binding Energy (B.E) Of The
Nucleus.
- To Disrupt A Stable Nucleus Into Its Constituent Protons And
Neutrons, The Minimum Energy Required Is The Binding Energy.
- The Magnitude Of The B.E. Of A Nucleus Determines Its Stability
Against Disintegration. If The B.E. Is Large, The Nucleus Is
Stable.
- A Nucleus Having The Least Possible Energy, Equal To The B.E.,
Is Said To Be In The Ground State.
- If The Nucleus Has An Energy E > Emin, It Is Said To Be In The
Excited
State.
- The Case E = 0 Corresponds To Dissociation Of The Nucleus Into
Its Constituent Nucleons.
- If M Is The Experimentally Determined Mass Of A Nuclide
Having Z Protons And N Neutrons,
B.E. = {(Zmp + Nmn)
- M } c^2
- If B.E. > 0, The Nucleus Is Stable And Energy Must Be Supplied
From Outside To Disrupt It Into Its Constituents. If B.E.< 0, The
Nucleus Is Unstabe And It Will Disintegrate By Itself.
- Example :
Let Us Illustrate The Calculation Of B.E. By Taking The Example
Of The Deuteron. A Deuteron Is Formed By A Proton And A
Neutron.
Mass Of Proton =
1.007276u.
Mass Of Neutron =
1.008665u.
Mass Of Proton +
Neutron In Free State = 2.015941u
Mass Of Deuteron
Nucleus = 2.013553u
Mass Defect = ∆m =
0.002388u
B.E. = 0.002388 ×
931 = 2.23 MeV ( 1u 931 MeV)
Neutron, 2.23 Mev Of Energy Is Liberated.
- Conversely, 2.23 Mev Must Be Supplied From An External
Source To Break A Deuteron Up Into A Proton And A Neutron.
- This Is Confirmed By Experiments That Show That A Gamma-
Ray Photon With A Minimum Energy Of 2.23 MeV Can Split A
Deuteron Into A
Free Neutron And A Free Proton Show In Fig
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