Presentation is loading. Please wait.

Presentation is loading. Please wait.

Nuclear Physics Chapter 25. 3 Li lithium 6.941 1 name symbol atomic number (# of p + ) average atomic mass electrons in outer energy level.

Similar presentations


Presentation on theme: "Nuclear Physics Chapter 25. 3 Li lithium 6.941 1 name symbol atomic number (# of p + ) average atomic mass electrons in outer energy level."— Presentation transcript:

1 Nuclear Physics Chapter 25

2 3 Li lithium 6.941 1 name symbol atomic number (# of p + ) average atomic mass electrons in outer energy level

3 Protons positive charge made of two up quarks and a down quark Electric charge +2/3 eElectric charge -1/3 e in the nucleus… Neutrons made of two down quarks and an up quark

4 Li 7 = proton = neutron number of nucleons (mass number) Li 6 an isotope of Li mass number talks about 1 atom atomic mass is an average of all atoms

5 Mass is usually expressed in amu or u 1 amu = 1/12 th the mass of a carbon 12 atom 1 amu = 1.66 x 10 -27 kg= mass of p + or n 0 an n 0 is actually about 5 x 10 -4 u more massive an e - has a mass of about 5 x 10 -4 u…hmmm

6 Mass can also be expressed in MeV/c 2 rest energy (E 0 ) = MeV Mega (1 x 10 6 ) electron volt (1.60 x 10 -19 J) 1 proton = 938.3 MeV/c 2

7 mass number atomic number (charge)

8 The electric force between protons is repulsive There must be a force stronger than the electric force holding the nucleus together

9 The Strong Force is the force that holds nucleons together. It is independent of charge, it binds neutrons and protons. It only works over very small distances (atomic).

10 10 -15 m r nucleus = (1.2x10 -15 m )A 1/3 where A is the atomic number

11 Stable “big” atoms have more neutrons than protons, because the electric force acts over greater distances.

12 Nucleons bound together in a nucleus have less mass (energy) than unbound nucleons E bind =  mc 2 binding energy is that energy that holds the nucleus together mass defect the difference in mass

13 Find the total binding energy of Al 27 if it has a mass of 26.981534 u. H 1 has a mass of 1.007825 u and the mass of a n 0 is 1.008665 u. There are 931.50 MeV/u. Al is atomic # 13, so it has 13(1.007825 u) of p + and e -, = 13.101725 u. Al has 14 n 0 = 14(1.008665 u) = 14.12131 u  m = (13.101725 + 14.12131) u – 26.981534 u =.241501 u x 931.50 MeV/u= 225 MeV

14 Nuclear Decay Alpha Decay: an  particle (He nucleus) escapes the strong force by quantum tunneling. The nucleus loses 2 p + and 2 n 0.

15 Beta decay: the weak force causes an up quark to change into a down quark, or vice versa. This causes a proton to change into a neutron or vice versa. neutronprotonelectron  - antineutrino

16 protonneutronpositron  + neutrino A thallium 208 nucleus emits a  - particle (an electron). Write the equation.

17 Gamma Decay: a nucleon can be in an excited state, needing to get rid of energy to reach a stable energy level. The nucleon emits a photon, but the structure of the nucleus stays the same.


Download ppt "Nuclear Physics Chapter 25. 3 Li lithium 6.941 1 name symbol atomic number (# of p + ) average atomic mass electrons in outer energy level."

Similar presentations


Ads by Google