IIT JEE Physics Formula Revision 46. Nucleus

1. R = R₀A^(1/3) .. (46.1)

where R₀ = 1.1*10^-15 m ≈ 1.1 fm and A is the mass number

2. B = [Zm{₁¹H} +Nm_n
– m{ _Z ^Z+n}]c²

Where
B = binding energy of the nucleus

m{₁¹H} is the mass of a hydrogen atom

m{ _Z ^Z+n} is the mass of an atom with Z protons and N neutrons


3. Binding energy per nucleon = B/A = a₁ - a₂/(A^1/3) - a₃Z(Z-1)/ (A^4/3)


4. Mass excess =
(mass of atom – A’)c²

5. Mass excess = 931(m-A)MeV

6. Alpha decay process is represented by

_Z^AX --> _Z-2^A-4Y + ₂⁴He

7. q value of the process

Q = [m{_Z^AX} – m{_Z-2^A-4Y} – m{₂⁴He}]c²

8. Beta decay process

N --> p + e + antineutrino

9. Beta decay proces
_Z^AX --> _Z+1^AY +e + antineutrino

e is also shown as beta minus.

10. kinetic energy Q available ot the product particles is

Q = [m(_Z^AX – m{_Z+1^AY}]c²

11. Proton conversion process – Beta plus decay

P --> n + e⁺ + v (neutrino)



12.
_Z^AX --> _Z-1^AY + e⁺ + v (neutrino)

13.Q value of the decay

Q = {m(_Z^AX) – m(_Z-1^AY) – 2m_e]c²