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Section 2: Mass Defect and E=mc 2.  Since an atom is made of protons, neutrons, and electrons, you might expect the mass of the atom to be the same as.

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Presentation on theme: "Section 2: Mass Defect and E=mc 2.  Since an atom is made of protons, neutrons, and electrons, you might expect the mass of the atom to be the same as."— Presentation transcript:

1 Section 2: Mass Defect and E=mc 2

2  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  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  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). E=mc 2 Energy (Unit = J) Δ Mass in Kg (mass of products– mass of reactants) Constant (3.00 x 10 8 m/s) Speed of light in a vacuum

5  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 1010 142 56 92 36 1010

6  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  Step 3: Solve for E using mass (kg) and constant (3.00 x 10 8 m/s) E=mc 2 E= (1.4447x10 -29 kg)(3.00x10 8 m/s) 2 E= (1.4447x10 -29 kg) 9.00x10 16 m 2 /s 2 E= 1.30023 x 10 -12 kg x m 2 /s 2 = 1.30 x 10 -12 J

8  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 1010 140 55 92 37 1010

9 E=mc 2 E= (3.3066x10 -27 kg)(3.00x10 8 m/s) 2 E= (3.3066x10 -27 kg) 9.00x10 16 m/s E= 2.9754x10 -10 kg x m 2 /s 2 = 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  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 2121 3131 4242 1010

11 E=mc 2 E= (1.4447x10 -29 kg)(3.00x10 8 m/s) 2 E= (1.4447x10 -29 kg) 9.00x10 16 m 2 /s 2 E= 1.30023 x 10 -12 kg x m 2 /s 2 = 1.30 x 10 -12 J The two hydrogen isotopes will release 1.30 x 10 -12 J of energy when they fuse together!


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