Wavelengths in this series can be represented by the

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wavelengths in this series can be represented by the formula 2 (75.00 nm)( 4), 3, 4, 5, . . . n n λ = = An alternative formula that starts from n = 1 is ( 4)(75.00 nm), 1, 2, 3, . . . n n n λ = + =
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38.38. Solve: (a) V atom = 3 atom 4 3 r π = 9.0 × 10 31 m 3 . Since the atomic mass of aluminum is 27 u, the mass per atom is m = 27 × (1.661 × 10 27 kg) = 4.5 × 10 26 kg Therefore, the average density of an aluminum atom is 26 3 atom 31 3 4.5 10 kg 50,000 kg/m 9.0 10 m ρ × = = × (b) The average density of an aluminum atom is found in part (a) to be larger than the density of solid aluminum ( ρ Al = 2700 kg/m 3 ). The volume per atom in the solid is 26 29 3 29 3 3 4.5 10 kg 1.67 10 m 1.7 10 m 2700 kg/m × = × × So, using 3 29 3 sphere 4 1.67 10 m , 3 r π = × we get r sphere = 1.6 × 10 10 m . That is, the average spacing between atoms is 2 r sphere = 3.2 × 10 10 m, which is larger than the diameter of the aluminum atom by a factor of 2.7. (c) The atomic mass is almost entirely in the nucleus, so the density of the nucleus is ( ) 26 17 3 nucleus 3 15 4 3 4.5 10 kg 1.7 10 kg/m 4 10 m π ρ × = = × × Compared to the density of solid aluminum, which is 2700 kg/m 3 , the nuclear density is approximately 10 14 times larger.
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38.39. Model: The nucleus of an atom contains Z protons and A Z neutrons. Solve: (a) The values of Z , A , and Z/A for nuclei with Z = 1, 5, 10, 15, 90 are given in the following table. Z 1 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 A 1 10.8 20 31 40 55 65 80 91 103 119 133 144 159 173 186 201 210 232 Z/A 1 0.46 0.5 0.48 0.5 0.45 0.46 0.44 0.44 0.44 0.42 0.41 0.42 0.41 0.40 0.40 0.40 0.40 0.39 A graph of Z/A as a function of Z is shown below. (b) The / Z A values are in the range 0.45–0.50 for values of Z up to approximately 45. For Z in the range 65 to 90, / Z A drops down to approximately 0.40. In short, / Z A decreases from around 0.50 to 0.40 with an increase in Z . (c) What we are plotting is / Z Z N + as a function of Z . With increasing Z , / Z A decreases because the number of neutrons in the nuclei increases more rapidly than Z .
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38.40. Model: The nucleus of an atom is very small and it contains protons and neutrons. Solve: (a) The repulsive electric force between two protons in the nucleus is ( ) ( )( ) ( ) 2 9 2 2 19 2 E 2 2 15 0 8.99 10 N m /C 1.60 10 C 1 58 N 4 2.0 fm 2.0 10 m e F πε × × = = = × (b) The attractive gravitational force between two protons in the nucleus is ( ) ( )( ) ( ) 2 11 2 2 27 2 35 G 2 2 15 6.67 10 N m /kg 1.67 10 kg 4.7 10 N 2.0 fm 2.0 10 m Gm F × × = = = × × Because F G << F E , gravitational force could not be the force to hold two protons together. (c) The nuclear force must be very strong to overcome F E and it must be independent of charge because both protons and neutrons are held in the nucleus very tightly. Furthermore, nuclear force is a very short range force since it is not felt outside the nucleus.
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38.41. Model: Assume the nucleus is at rest. Use the conservation of energy equation. Visualize: Solve: The energy conservation equation K f + U f = K i + U i is ( )( ) ( ) ( ) ( ) ( ) 0 2 9 2 2 19 19 6 15 2 1 0 J 6.24 MeV 0 J 4 6 fm 9.0 10 N m /kg 2 1.60 10 C 1.60 10 J 6.24 10 eV 6.0 10 m 1 eV e Ze Z πε + = + × × × = × × × Solving for Z gives Z = 13. The element is aluminum.
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38.42.
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