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# Proton A nuclear particle having a positive charge equal to that of the electron and a mass more than 1800 times that of

Proton A nuclear particle having a positive charge equal to that of the electron and a mass more than 1800 times that of the electron The number of protons in an atom is called the atomic number, Z.

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## Proton A nuclear particle having a positive charge equal to that of the electron and a mass more than 1800 times that of

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1. Proton A nuclear particle having a positive charge equal to that of the electron and a mass more than 1800 times that of the electron The number of protons in an atom is called the atomic number,Z. An element is a substance whose atoms have the same number of protons and thus the same atomic number, Z.

2. Neutron A nuclear particle having a mass almost equal to that of the proton but no electrical charge The mass number,A, is the total number of protons and neutrons in the nucleus. Isotopes are atoms whose nuclei have the same atomic number (number of protons) but different numbers of neutrons (mass number).

3. Fig. 2.13

4. Nuclide • An atom characterized by a certain atomic number, Z, and mass number, A. • Nuclide symbol • Examples:

5. Atomic Definitions I: Symbols, Isotopes,Numbers A X The Nuclear Symbol of the Atom, or Isotope Z X = Atomic symbol of the element, or element symbol A = The Mass number; A = Z + N Z = The Atomic Number, the Number of Protons in the Nucleus N = The Number of Neutrons in the Nucleus Isotopes = atoms of an element with the same number of protons, but different numbers of Neutrons in the Nucleus

6. Write the nuclide symbol for the atom that has 19 protons and 20 neutrons. Atomic number: Z = 19 The element is potassium, K. Mass number: A = 19 + 20 = 39 The nuclide symbol is

7. Depicting the Atom Fig. 2.14

8. Neutral ATOMS • 51 Cr = P+ (24), e- (24), • N (27) • 239 Pu = P+(94), e- (94), • N (145) • 15 N = P+(7), e-(7), N(8) • 56 Fe = P+(26), e-(26), • N (50) • 235 U =P+(92), e-(92), • N (143)

9. Atomic Mass The average atomic mass for the naturally occurring element expressed in atomic mass units (amu) (Note that this is a weighted average of the relative abundances and masses of the isotopes.) Atomic Mass Unit Equal to exactly one-twelfth the mass of a carbon-12 atom

10. Isotopes of Hydrogen 11H 1 Proton 0 Neutrons 99.985 % 1.00782503 amu 21H (D) 1 Proton 1 Neutron 0.015 % 2.01410178 amu 31H (T) 1 Proton 2 Neutrons -------- ---------- The average mass of Hydrogen is 1.008 amu • 3H is Radioactive with a half life of 12 years. • H2O Normal water “light water “ • mass = 18.0 amu , BP = 100.000000C • D2O Heavy water • mass = 20.0 amu , BP = 101.42 0C

11. Figure 2.9: A representation of two isotopes of carbon.

12. Figure 2.11: Diagram of a simple mass spectrometer.

13. Figure 2.12: The mass spectrum of neon.

14. Element #8 : Oxygen, Isotopes • 168O 8 Protons 8 Neutrons • 99.759% 15.99491462 amu • 178O 8 Protons 9 Neutrons • 0.037% 16.9997341 amu • 188O8 Protons 10 Neutrons • 0.204 % 17.999160 amu

15. Calculating the “Average” Atomic Mass of an Element Problem: Calculate the average atomic mass of Magnesium! Magnesium Has three stable isotopes, 24Mg ( 78.7%); 25Mg (10.2%); 26Mg (11.1%). 24Mg (78.7%) 23.98504 amu x 0.787 = 18.876226 amu 25Mg (10.2%) 24.98584 amux 0.102 = 2.548556 amu 26Mg (11.1%) 25.98636 amu x 0.111 = 2.884486 amu 24.309268 amu With Significant Digits = 24.3 amu

16. Problem: Calculate the abundance of the two Bromine isotopes: 79Br = 78.918336 g/mol and 81Br = 80.91629 g/mol , given that the average mass of Bromine is 79.904 g/mol. Plan: Let the abundance of 79Br = X and of 81Br = Y and X + Y = 1.0 Solution: X(78.918336) + Y(80.91629) = 79.904 X + Y = 1.00 therefore X = 1.00 - Y (1.00 - Y)(78.918336) + Y(80.91629) = 79.904 78.918336 - 78.918336 Y + 80.91629 Y = 79.904 1.997954 Y = 0.985664 or Y = 0.4933 X = 1.00 - Y = 1.00 - 0.4933 = 0.5067 %X = % 79Br = 0.5067 x 100% = 50.67% = 79Br %Y = % 81Br = 0.4933 x 100% = 49.33% = 81Br

17. An element has four naturally occurring isotopes. The mass and percentage of each isotope are as follows: What is the atomic weight and name of the element?

18. To find the portion of the atomic weight due to each isotope, multiply the fraction by the mass of the isotope. The atomic weight is the sum of these products. The atomic weight is 207 amu; the element is lead.

19. 2. (8 points) Atoms X, Y, Z, and R have the following nuclear compositions: Which two are isotopes of each other? Circle them and explain.

20. 6. (11 points) Fill in the blank spaces and write out all the symbols in the left hand column in full, in the form (i.e. include the appropriate values of Z and A as well as the correct symbol X and the charge, if any).

21. (15 points) Lithium forms compounds which are used in dry cells and storage batteries and in high-temperature lubricants. It has two naturally occurring isotopes, 6Li (isotopic mass = 6.015121 amu) and 7Li (isotopic mass = 7.016003 amu). Lithium has an atomic mass of 6.9409 amu. What is the percent abundance of lithium-6?

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