# Model for a neutral atom the number of electrons is

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Model: For a neutral atom, the number of electrons is the same as the number of protons, which is the atomic number Z . An atom’s mass number is A = Z + N , where N is the number of neutrons. Solve: (a) For a 207 Pb atom, Z = 82. So, N = 207 – 82 = 125. The neutral 207 Pb atom contains 82 electrons, 82 protons, and 125 neutrons. (b) The electric potential at the surface of a lead nucleus is ( ) ( ) ( ) 9 2 2 19 7 15 0 9.0 10 N m /C 82 1.60 10 C 1 1.66 10 V 4 7.1 10 m Q V r πε × × = = = × × The electric field strength at the surface of a lead nucleus is ( ) ( ) ( ) ( ) 9 2 2 19 21 2 2 15 0 9.0 10 N m /C 82 1.60 10 C 1 2.34 10 V/m 4 7.1 10 m Q E r πε × × = = = × ×

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38.19. Model: Use Equation 38.11 which is the Balmer formula. Visualize: Please refer to Figure 38.22. Solve: (a) The wavelengths in the hydrogen emission spectrum are 656.6 nm, 486.3 nm, 434.2 nm, and 410.3 nm. The formula for the Balmer series can be written 2 2 1 1 91.18 nm m n λ = where m = 1, 2, 3, . . . and n = m + 1, m + 2, . . . For the first wavelength, 2 2 2 2 2 2 1 1 91.18 nm 0.1389 0.1389 656.5 nm n m m n n m = = = This equation is satisfied when m = 2 and n = 3. For the second wavelength (486.3 nm) the equation is satisfied for m = 2 and n = 4. Likewise, for the next two wavelengths m = 2 and n = 5 and 6. (b) The fifth line in the spectrum will correspond to m = 2 and n = 7. Its wavelength is ( ) ( ) ( ) 2 2 1 1 2 7 91.18 nm 196 91.18 nm 397.1 nm 45 λ = = =
38.20. Model: Use Equation 38.11 which is the Balmer formula. Visualize: Please refer to Figure 38.22. Solve: (a) The wavelengths in the hydrogen emission spectrum are 656.6 nm, 486.3 nm, 434.2 nm, and 410.3 nm. The formula for the Balmer series can be written 2 2 1 1 91.18 nm m n λ = where m = 1, 2, 3, . . . and n = m + 1, m + 2, . . . For the first wavelength, 2 2 2 2 2 2 1 1 91.18 nm 0.1389 0.1389 656.5 nm n m m n n m = = = This equation is satisfied when m = 2 and n = 3. For the second wavelength (486.3 nm) the equation is satisfied for m = 2 and n = 4. Likewise, for the next two wavelengths m = 2 and n = 5 and 6. (b) From part (a), the lines in the spectrum in Figure 38.22 (b) are represented by the Balmer formula when m = 2. The limiting wavelength of the series occurs when 2 , which means 1 0. n n → ∞ In this limit, ( ) ( ) ( ) ( ) 2 2 2 1 1 1 2 2 91.18 nm 91.18 nm 4 91.18 nm 364.7 nm n λ = = =

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38.21. Model: Use Equation 38.11 with m = 1 and n = 2, 3, 4, and 5. This series of spectral lines is called the Lyman series. Solve: The formula for the Lyman series simplifies to ( ) 2 91.18 nm 1 1 n λ = where n = 2, 3, 4 and 5 For n = 2, ( ) ( ) 1 1 4 91.18 nm 1 121.6 nm. λ = = For n = 3, ( ) ( ) 1 1 9 91.18 nm 1 102.6 nm. λ = = Likewise, for n = 4 and n = 5, 97.3 nm λ = and 95.0 nm, λ = respectively.
38.22. Model: Use Equation 38.9 which is the Balmer formula. Solve: (a) Equation 38.9 can be written as 2 2 2 2 2 2 1 1 91.18 nm n m m n n m λ = = where m = 1, 2, 3, . . . and n = m + 1, m + 2, . . . For λ = 102.6 nm, 2 2 2 2 91.18 nm 0.8887 102.6 nm n m n m = = This equation is satisfied by m = 1 and n = 3. For λ = 1876 nm, 2 2 2 2 91.18 nm 0.0486 1876 nm n m n m = = This equation is satisfied by m = 3 and n = 4. (b) The ultraviolet wavelengths are <400 nm. The infrared wavelengths are >700 nm. So, λ = 102.6 nm is in the ultraviolet region and λ = 1876 nm is in the infrared region.

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38.23.
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