ch4 - Chapter 4 ! The Nuclear Atom 4-1. The Lyman series...

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75 Chapter 4 ! The Nuclear Atom 4-1. The Lyman series ends on m = 1, the Balmer series on m = 2, and the Paschen series on m = 3. The series limits all have n = 4 , so 4-2.
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Chapter 4 ! The Nuclear Atom 76 4-3. No, this is not a hydrogen Lyman series transition because n is not an integer. 4-4. For the Brackett series m = 4 and the first four (i.e., longest wavelength) lines have n = 5, 6, 7, and 8. .Similarly, These lines are all in the infrared.
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Chapter 4 ! The Nuclear Atom 77 4-5. None of these lines are in the Paschen series, whose limit is 820.4 nm (see Problem 4-1) and whose first line is given by: . Also, none are in the Brackett series, whose longest wavelength line is 4052 nm (see Problem 4-4). The Pfund series has m = 5. Its first three (i.e., longest wavelength) lines have n = 6, 7, a nd 8. . Similarly, Thus, the line at 4103 nm is not a hydrogen spectral line. 4-6. (a) For Au, (see Example 4-2) and for this foil . (b) For ,
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Chapter 4 ! The Nuclear Atom 78 (Problem 4-6 continued) and For , and Therefore, 4-7. (from Equation 4-6), where A is the product of the two quantities in parentheses in Equation 4-6. (a) (b) 4-8. 4-9.
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Chapter 4 ! The Nuclear Atom 79 (Problem 4-9 continued) 4-10. 4-11. 10 4 atomic layers is not enough to produce a deflection of 10°, assuming 1 collision/ layer. 4-12. (a) For (refer to Problem 4-6). Because For ,
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Chapter 4 ! The Nuclear Atom 80 (Problem 4-12 continued) Because (b) (c) For 2 = 75°, For 2 = 90°, 4-13. (a) (b) 4-14.
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Chapter 4 ! The Nuclear Atom 81 (Problem 4-14 continued) 4-15. None of these are in the visible; all are in the ultraviolet.
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Chapter 4 ! The Nuclear Atom 82 4-16. (Equation 4-17) (from Equation 4-17) This would not be detectable. or The orbit radius r would still be 4-17. (a) is in the visible region of the spectrum, so this is a transition ending on (see Figure 4-16). (b) This line is in the Balmer series.
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Chapter 4 ! The Nuclear Atom 83 4-18. The number of revolutions N in 10 ! 8 s is: The radius of the orbit is given by so the circumference of the orbit C = 2 B r is The electron's speed in the orbit is given by Therefore, In the planetary analogy of Earth revolving around the sun, this corresponds to 3.7 million "years". 4-19. (a) (b) (c )T he shortest wavelength in the Lyman series is the series limit (n i = 4 , n f = 1). The photon energy is equal in magnitude to the ground state energy ! E : .
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ch4 - Chapter 4 ! The Nuclear Atom 4-1. The Lyman series...

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