Presentation is loading. Please wait.

Presentation is loading. Please wait.

Basic Nuclear Chemistry. Line vs. Continuous Spectra.

Similar presentations


Presentation on theme: "Basic Nuclear Chemistry. Line vs. Continuous Spectra."— Presentation transcript:

1 Basic Nuclear Chemistry

2 Line vs. Continuous Spectra

3 Continuous Spectrum all colors (wavelengths) are present prism separates different wavelengths white light photographic plate

4 Line Spectrum

5 7.3 Important Information: Line location Line intensity

6 Nuclear Atom electron (e - ) 9.10939 x 10 -28 g charge = -1 proton(p + ) 1.672623 x 10 -24 g charge = +1 neutron (n o ) 1.674929 x 10 -24 g no charge

7 A 19 X F Z 9 Isotopic Symbol element symbol atomic number Z = # protons mass number A = Z + #neutrons

8 Isotopes 12 1314 C C C 6 6 6 These are isotopes of carbon (same Z, different A). Only 14 C is radioactive (unstable).

9 The Stable Isotopes

10 Nuclear Reactions Fusion - joining of 2 or more nuclei Fission - splitting of a nucleus radioactive decay - emission of particles and/or radiation from the nucleus

11 1p1p 1 1H1H 1 or proton 1n1n 0 neutron 0e0e 00 or electron (beta-) 0e0e +1 00 or positron (beta+) 4 He 2 44 2 or  particle Particles Involved in Nuclear Rxns …and  (gamma) radiation (not a particle)

12 n/p too large n/p too small beta decay positron emission electron capture

13 Geiger Counter

14 Radioactive Decay and Shielding (Problem #6)

15 Balancing Nuclear Equations 1.Conserve mass number (A) 1n1n 0 U 235 92 + Cs 138 55 Rb 96 37 1n1n 0 ++ 2 235 + 1 = 138 + 96 + 2x1 2.Conserve atomic number (Z) 1n1n 0 U 235 92 + Cs 138 55 Rb 96 37 1n1n 0 ++ 2 92 + 0 = 55 + 37 + 2x0

16 Problem #7: 212 Po decays by alpha emission. Write the balanced nuclear equation for the decay of 212 Po. 4 He 2  = 212 = 4 + A; A = 208 84 = 2 + Z; Z = 82 212 Po 4 He + 208 Pb 84 282 212 Po 4 He + A X 84 2Z

17 Radioisotopes in Medicine 1/3 of all hospital patients undergo nuclear medicine procedures. Brain images with 123 I-labeled compound

18  24 Na, t 1/2 = 14.8 hr,  emitter, blood-flow tracer  131 I, t 1/2 = 8 hr,  emitter, thyroid gland activity  123 I, t 1/2 = 13.3 hr,  ray emitter, brain imaging  18 F, t 1/2 = 1.8 hr,   emitter, positron emission tomography  99m Tc, t 1/2 = 6 hr,  ray emitter, imaging agent Half Lives of Medical Isotopes t 1/2 = time for half the nuclide to decay These nuclides are chosen because they have short half lives. Why?


Download ppt "Basic Nuclear Chemistry. Line vs. Continuous Spectra."

Similar presentations


Ads by Google