The document summarizes key concepts regarding nuclear forces and nuclear structure: 1) Nuclear forces are very strong over short ranges (~femtometers) and act independently of charge between nucleons. The force is attractive at longer ranges but repulsive at very short ranges. 2) Basic properties of nuclei include that atomic mass (A) equals protons + neutrons, atomic number (Z) equals protons, and nuclear density is constant regardless of mass number. 3) Binding energy explains nuclear stability - breaking nuclei requires energy equal to the binding energy per nucleon, which is highest for mid-sized nuclei like iron. 4) Radioactive decay processes like alpha, beta, and gamma decay occur to reach

1. Nuclei

2. Nuclear Forces
Short Ranged Force ( ~ femtometers 10-15)
In short range , they are very strong(even stronger than Coulomb and Gravitation) .
Other than this they are very weak.
Nuclear force does not depend on charge , Hence it is same for neutron-neutron ,
proton-proton and neutron-proton
 From Graph: r > r0 attractive
r<ro repulsive
R0
Strong force
region

3. Basic Facts
 A = Mass number = No of protons + Number of electrons
Z = Atomic Number = No of protons
Radius of Nucleus ∝ A1/3
Volume = (4/3) π R3 ∝ A
Density = Mass / Volume = Mass number /Volume = Constant
So Density of nucleus is independent of Mass number .
Density of nucleus is very high . As it was concluded in Rutherford’s Experiment

4. Mass Energy Equivalence
Mass is condensed form of Energy
Energy Equivalent for Mass : E = mc2
Energy Equivalent of 1 amu = 931 MeV. ~930
So , If mass is known in amu just multiply it by 931 to get energy equivalent .

5. Energy Concepts
To break things we need energy . Example to break a chemical bond energy is
required , To break an ice cube ,energy is required .
When things are formed on their own , energy is released .

6. Binding Energy
Since we are breaking , Energy should be added . Energy Equal to Binding Energy is
required to break nucleus . So Binding Energy Stability
Applying Energy Conservation
Energy of Nucleus + Binding Energy = Energy of nucleons
Mnucleusc2 + Binding Energy = Mnucleons c2
Mnucleons > Mnucleus .. This is called mass defect (खुल्ले का mass ज्यादा होता है)
So Binding Energy = (Mass Defect) c2 = (Mnucleons-Mnucleus) c2

7. Consider a nucleus having Atomic Number Z and Mass number A and Atomic mass M
Number of Proton = Z
Number of neutrons = A- Z
Mass of Nucleons = Z mproton + (A-Z ) mneutron
Energy Equivalent = (Z mproton + (A-Z ) mneutron) c2
Mass of Nucleus = M
Energy Equivalent = Mc2
Energy of Nucleus + Binding Energy = Energy of Nucleons
Mc2 + Binding Energy =(Z mproton + (A-Z ) mneutron) c2
But Binding Energy is not a good measure , Rather (Binding Energy / nucleon) is better

8. Example
Protons = 8 Neutrons = 16-8=8
But Actual Mass = Atomic Mass = ( Mass Defect !!)
Energy of Nucleus + Binding Energy = Energy of nucleons
(15.99493)x931 + Binding = (16.12744)X931
Binding Energy = 127.5 MeV No of nucleon =16
Binding energy/ nucleons = 127.5 / 16 =7.96 MeV

9. Binding Energy Curve
Curve is on Binding Energy /Nucleon
Most stable : Iron Group
30< A<70 almost constant , because nuclear force is
short ranged .
Binding Energy / nucleon is low for light as well as heavy elements

10. Interpretations of curve
Light elements can combine to form a bigger element So that
BE / nucleon increase . Hence they are suitable for Fusion
Similarly ,Heavy Elements can break into two to form elements
in mid . So they are suitable for fission
Middle Elements are stable , so they are not suitable for any nuclear reactions

11. Alpha Decay
Alpha Particle ( ) is emitted
Energy Calculations :
Initial Mass = Mx  Initial Energy = Mx c2
Final Mass = (My + MHe ) c2
If Initial Mass is higher  Initial energy is more . So , Energy will be released(spontaneous)
If Final Mass is higher  Final energy is more . So Energy will be absorbed .
Energy released is carried by alpha particle in form of kinetic energy
Qvalue = (initial mass – final mass ) c2 = (Mx – My-MHe ) c2
Qvalue> 0 for spontaneous (Reactions with negative Q will not occur)

12. Example
Uranium
Initial Mass = 238.05079 u
Final Mass = 234.04363 + 4.00260 = 238.04623 u
Initial Mass > Final Mass
Initial Energy > Final Energy  Energy will be Released  Spontaneous
Energy released = (238.05079 -238.04623 ) X 931
= 0.00456 X931 MeV = 4.35 MeV
1 u = 931 MeV

13. Beta - decay
Electron is emitted . Ve is antineutrino ( zero mass)
Neutron breaks into electron and proton
Explanation
LHS : Protons = Z Neutrons = A-Z Electrons = Z
RHS : Since One Neutron has converted into electron + proton
Protons = Z+1 Neutrons = (A-Z)-1 Electrons = Z+1
So New Atomic number = Z+1 New Mass Number = (Z+1) +(A-Z)-1 = A !
Z Z +1 A unchanged

14. Energy Calculations
Initial Mass = Mx
Initial Energy = Mxc2
Final Mass = (My + Me)
Final Energy = (My + Me)c2
Qvalue = ( Initial Mass – Final Mass ) c2 = (Mx- My -Me)c2
Qvalue > 0  Spontaneous
This energy ( Q) is carried away as kinetic energy by the electron and
antineutrino.

15. Beta plus decay
Proton breaks into Neutron and positive electron
Explanation :
LHS : Protons : Z Neutrons : A-Z
Now one proton converts into positive electron and neutron
RHS : Proton : Z-1 Neutron : (A-Z+1)
New Atomic number : Z-1 New Mass Number = (Z-1) + (A-Z+1 ) = A !
Z  Z-1 A  Unchanged

16. Energy Calculations
Initial Mass = Mx
Initial Energy = Mxc2
Final Mass = (My + Me)
Final Energy = (My + Me)c2
Qvalue = ( Initial Mass – Final Mass ) c2 = (Mx- My -Me)c2
Qvalue > 0  Spontaneous
This energy(Q) is carried away as kinetic energy by the electron and neutrino.

17. Gamma Decay
Like electrons , nuclei also have energy states .
But energy gap is very high in case of nucleus . In case of electrons it is relatively small ( 13.6 eV,
10.2eV etc)
But in case of nucleus it is of the order of MeV ( Mega = 10 6 , as in megabytes)
When nucleus , jumps down from excited state to lower state energy is released in the form of
photons . Efinal- Einitial = h v

18. Example
Fact : Beta / alpha decay are generally accompanied by gamma decay
In this example , Cobalt first goes beta minus decay , In beta
Minus decay Z increases by 1 ( 2728)
Then it undergoes two gamma decays
First gamma : 1.17 Mev 1.17X106 = 1240/ λ λ =10-3 nm
Second gamma : 1.33 MeV

19. Radioactivity
First Order Reaction
N = Number of nuclei present
N0 initial number of nuclei
So number of nuclei will decrease with time .

20. Decay rate :
So ,
Putting value of N in above equation
R is also called activity of sample

21. Half life : Time at which number of nuclei present become half of initial nuclei
T1/2 depends only on decay constant .not on N
Average life : Life a nuclei is expected to live .
Like some humans die at early age , some at high age . The average comes in
middle which is called life expectancy .

22. Moderator – For slowing down Neutrons
Safety rods – to absorb neutrons