1 / 16

Notes 42 - Topic 7 - Atomic and Nuclear Physics

Notes 42 - Topic 7 - Atomic and Nuclear Physics - ------------------------------------------------------------------------------------------

nizana
Download Presentation

Notes 42 - Topic 7 - Atomic and Nuclear Physics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Notes 42 - Topic 7 - Atomic and Nuclear Physics ------------------------------------------------------------------------------------------- 7.3.1 Artificial Transmutation - an artificially induced nuclear reaction, resulting in a change of atomic number, caused by the bombardment of a nucleus with subatomic particles or small nuclei; Al-27 bombarded by alpha yields P-30 and a neutron (show equation in NB)

  2. 7.3.2 Nuclear Reaction Equations In 1919, Rutherford bombarded nitrogen gas with alpha particles from radioactive Bismuth-214. He discovered that radioactive Oxygen-17 was produced along with a massive, + charged particle. In one experiment, he discovered artificial transmutation and the proton!! (show equations in NB) N-14 bombarded by alpha yields O-17 and a proton... Other examples of artificial transmutation: N-14 bombarded by a neutron yields C-14 and a proton...

  3. O-16 bombarded by a deuterium yields N-14 and an alpha particle: Li-7 bombarded by a proton yields two alpha particles: O-16 bombarded by a neutron yields a deuterium and what else? • The bombardment of nuclei by all particles except neutrons requires the use of a particle accelerator;

  4. 7.3.3 Unified Mass Unit (u)* or Atomic Mass Unit (amu)** • Defined as 1/12 the mass of a C-12 atom (12 p’s + 12 n’s + 12 e’s); 1.00 mol C-12 = 12.0 g = 0.0120 kg 1.00 mol C-12 = 6.02 x 1023 atoms 1 atom C-12 = (0.0120 kg / 6.02 x 1023 atoms) = 1.99 x 10-26 kg atom-1 So... u = (1.99 x 10-26 kg / 12) = 1.66 x 10-27 kg***; 1 u = 1 amu = 1.66 x 10-27 kg examples: me = 0.000549 u mp = 1.007277 u*** mn = 1.008665 u*** mH = 1.007825 u*** ***SigFig’s are NOT followed closely in nuclear calculations

  5. 7.3.4 Einstein’s Mass-Energy Equivalence...E = mc2 - part of the Special Theory of Relativity published in 1905 which states that energy and mass are different forms of the same thing. • Energy can be converted to mass and mass can be converted to energy according to E = mc2, where E is energy in joules, m is mass in kg and c is the speed of light (3.0 x 108 ms-1). Sample Problem: How much energy is released when 1.00 kg of U-235 is changed into energy by nuclear fission in a nuclear power plant? Given: Unknown: Equation: (Note: This is approximately the same amount of energy as that released by burning 15,000 kg of oil or 20,000 kg of coal.)

  6. 7.3.5 Mass Defect and Binding Energy; 7.3.6 Solve problems involving mass defect and binding energy; Every nucleus larger than hydrogen has two or more protons. Each proton is positively charged and repels the other proton(s). What keeps a nucleus greater than hydrogen from “flying apart?”

  7. 1. Determine the mass of a helium atom in u’s: One helium atom has 2 e’s, 2 p’s, and 2 n’s 2 x (0.000549 u) 2 x (1.007277 u) +2 x (1.008665 u) 4.032982 u = mHe 2. But, the actual measured mHe = 4.002602 u; 4.032982 u -4.002602 u 0.030380 u There is a difference between the actual and the expected masses of 0.03038 u - this is called the mass defect;

  8. • A New Unit for Mass u = 1.661 x 10-27 kg; c = 2.998 x 108 ms-1; E = mc2 = (1.661 x 10-27 kg) (8.988 x 1016 m2s-2) = 1.49291 x 10-10 J; 1.000 eV = 1.602 x 10-19 J; So... E = (1.493 x 10-10 J / 1.602 x 10-19 J eV-1) = 9.315 x 108 eV; E = 9.315 x 108 eV = 931.5 MeV; Since E = mc2, m = E / c2; Finally, the mass of u... 1u = 931.5 MeVc-2; • Summary...look at the masses of the p, n, & e in MeVc-2 listed in the IB Physics Data Booklet!!

  9. 4. Experimentally, the amount of energy needed to break a helium nucleus apart, ie., the binding energy of the nucleus, is 4.54 x 10-12 J or 27.2 MeV; 5. The mass defect of the nucleus must be converted into the binding energy of the strong nuclear force needed to hold the nucleus together. 6. We compare stability of nuclei by comparing their binding energy per nucleon.

  10. 3. How much energy does the mass defect represent? Given: Unknown: Energy (in Joules) and Energy (in MeV) Equations: 0.03038 u - this is called the mass defect; 1 u of mass = 931.5 MeV/c2 of mass 1 u of mass = 931.5 MeV of energy 1 u of mass = 1.49291 x 10-10 J of energy

  11. Determine the binding energy per nucleon for Helium (in MeV). Given: Unknown: BE/nucleon Equation: BE =28.29897 MeV

  12. Nuclear Binding Energy Practice--This will be turned in. examples: me = 0.000549 u mp = 1.007277 u*** mn = 1.008665 u*** 1. Pick five neutral isotopes that you like (not necessarily stable). 2. For each one, calculate the mass defect in u AND MeV/c2. 3. For each one, calculate the binding energy in MeV. 4. For each one, calculate the binding energy per nucleon in MeV. 5. What's the most stable isotope that you picked initially?

  13. 7.3.6 Draw by hand, and annotate, on a full page of graph paper, a graph of binding energy per nucleon as a function of nucleon number, and apply it to predict nuclear energy changes for fission and fusion; ..

  14. 238 92 1 0 90 38 146 56 1 0 n + U Sr Xe + + 3 n How does this nuclear reaction yield energy?

More Related