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P214hw_sol_10

# P214hw_sol_10 - Solution to P214 HW 10 Prepared by Seong...

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Solution to P214 HW # 10 Prepared by Seong Chan Park 1. (38.1) λ = 520nm. c = λf f = c λ = 3 . 00 × 10 8 m / s 520 × 10 - 9 m = 5 . 77 × 10 14 Hz . p = h/λ p = 6 . 63 × 10 - 34 J · s 520 × 10 - 9 m = 1 . 27 × 10 - 27 J · s / m . E = hf E = 6 . 63 × 10 - 34 J · s × 5 . 77 × 10 14 Hz = 3 . 83 × 10 - 19 J = 2 . 39eV . where we have used (1eV = 1 . 602 × 10 - 19 J) as the conversion factor. 2. (38.2) λ = 652nm , Δ t = 20 . 0ms , ¯ P = 0 . 600W. ( a ) E = ¯ P Δ t = 0 . 600W × 20 . 0 × 10 - 3 s = 1 . 20 × 10 - 2 J = 7 . 49 × 10 16 eV ( b ) E = hf = h c λ = 6 . 63 × 10 - 34 J · s × 3 . 00 × 10 8 m / s 652 × 10 - 9 m = 3 . 05 × 10 - 19 J = 1 . 90eV ( c ) n = 7 . 49 × 10 16 1 . 90 = 3 . 94 × 10 16 . 3. (38.17) Balmer’s formula for transition from n -th state is: 1 λ n 2 = R ( 1 2 2 - 1 n 2 ) , and H γ mode corresponds to 5 2 transition (see Fig. 38.8). ( a ) 1 λ = R (1 / 2 2 - 1 / 5 2 ) . λ = 1 R × 100 21 = 4 . 33 × 10 - 7 m . ( b ) f = c λ = 3 . 00 × 10 8 m / s 4 . 33 × 10 - 7 m = 6 . 93 × 10 14 Hz . ( c ) E = hf = 6 . 63 × 10 - 34 Js × 6 . 93 × 10 19 Hz = 2 . 87eV . 1

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4. (38.23) The ion, Be 3+ , behaves very much like a hydrogen atom except that the nuclear charge is four times as great. Accordingly, the Coulomb force is four times stronger.
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