Thursday, February 7, 2008

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} +Nmn
– 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/(A1/3) - a3Z(Z-1)/ (A4/3)

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

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

6. Alpha decay process is represented by

ZAX --> Z-2A-4Y + 24He

7. q value of the process

Q = [m{ZAX} – m{Z-2A-4Y} – m{24He}]c²

8. Beta decay process

N --> p + e + antineutrino

9. Beta decay proces
ZAX --> Z+1AY +e + antineutrino

e is also shown as beta minus.

10. kinetic energy Q available ot the product particles is

Q = [m(ZAX – m{Z+1AY}]c²

11. Proton conversion process – Beta plus decay

P --> n + e+ + v (neutrino)

12.
ZAX --> Z-1AY + e+ + v (neutrino)

13.Q value of the decay

Q = {m(ZAX) – m(Z-1AY) – 2me]c²