1 / 11

Unit 14: Nuclear Chemistry

Unit 14: Nuclear Chemistry. Section 2: Mass Defect and E=mc 2. Mass Defect. Since an atom is made of protons, neutrons, and electrons, you might expect the mass of the atom to be the same as the mass of all of these combined. However, this is not always the case. Lets look at He...

blaine
Télécharger la présentation

Unit 14: Nuclear Chemistry

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. Unit 14: Nuclear Chemistry Section 2: Mass Defect and E=mc2

  2. Mass Defect • Since an atom is made of protons, neutrons, and electrons, you might expect the mass of the atom to be the same as the mass of all of these combined. • However, this is not always the case. Lets look at He... 2 Protons: (2 x 1.007276 amu) = 2.014552 amu 2 Neutrons: (2 x 1.008665 amu) = 2.017330 amu 2 Electrons: (2 x 0.0005486 amu) = 0.001097 amu Total combined = 4.032979 amu

  3. Mass Defect • The actual atomic mass of Helium has been measured to be 4.002602 amu… • The difference between the mass of an atom and the sum of the masses of its particles is called the mass defect. • Sooooooooo where did the rest of the mass go?! • According to Einstein, it was converted directly into energy! 

  4. Einstein’s Equation • Albert Einstein developed an equation to measure the amount of energy created when the nucleus is created from loose particles (fusion) or split apart (fission). Δ Mass in Kg (mass of products– mass of reactants) Energy (Unit = J) E=mc2 Constant (3.00 x 108 m/s) Speed of light in a vacuum

  5. E=mc2 • Step 1: Calculate the mass of the reactants and products… • U + n  Ba + Kr + 2 n • Mass of a neutron = 1.0087 amu • Reactants: U (235) + n (1.0087) = 236.0087 amu • Products: Ba (142) + Kr (92) + 2 n (1.0087 x 2) = 236.0174 amu • Difference: 236.0174 – 236.0087 = 0.0087 amu 235 92 1 0 142 56 92 36 1 0

  6. E=mc2 • Step 2: Convert mass to Kilograms by multiplying difference in mass (amu) by 1.6605 x 10-27 kg. 0.0087 amu x 1.6605 x 10-27 kg = 1.4447 x 10-29 kg Now this number can be plugged into the equation…

  7. E=mc2 • Step 3: Solve for E using mass (kg) and constant (3.00 x 108 m/s) E=mc2 E= (1.4447x10-29 kg)(3.00x108 m/s)2 E= (1.4447x10-29 kg) 9.00x1016 m2/s2 E= 1.30023 x 10-12 kg x m2/s2 = 1.30 x 10-12 J

  8. Sample Problem 1 • Calculate the energy released when we split U-235 into Cs and Rb. U + n  Cs+ Rb+ 2 n • 235 + 1.0087 = 236.0087 amu (Reac) • 140 + 92 + 2.0147 = 234.0174 amu (Prod) • Difference: 234.0174-236.0087=1.9913 amu x 1.6605x10-27 kg = 3.3066x10-27 kg 140 55 92 37 235 92 1 0 140 55 1 0 92 37

  9. Sample Problem 1 (continued) E=mc2 E= (3.3066x10-27 kg)(3.00x108 m/s)2 E= (3.3066x10-27 kg) 9.00x1016 m/s E= 2.9754x10-10 kg x m2/s2 = 2.98 x 10-10 J Uranium will release 2.98 x 10-10 J of energy when it splits into Cs and Rb!

  10. Sample Problem 2 • Calculate the energy released when two Hydrogen isotopes fuse. H + H  He + n • 2 + 3 = 5 amu (Reactants) • 4 + 1.0087 = 5.0087 amu (Products) • Difference: 5.0087-5 = 0.0087 amu • Convert to Kg: 0.0087 amu x 1.6605x10-27 kg = 1.4447 x 10-29 kg 3 1 4 2 2 1 1 0

  11. Sample Problem 2 (continued) E=mc2 E= (1.4447x10-29 kg)(3.00x108 m/s)2 E= (1.4447x10-29 kg) 9.00x1016 m2/s2 E= 1.30023 x 10-12 kg x m2/s2 = 1.30 x 10-12 J The two hydrogen isotopes will release 1.30 x 10-12 J of energy when they fuse together!

More Related