p40_044

# p40_044 - 44(a The calculation is shown in Sample Problem...

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Unformatted text preview: 44. (a) The calculation is shown in Sample Problem 40-6. The diﬀerence in the values obtained in parts (a) and (b) of that Sample Problem is 122 nm − 91.4 nm ≈ 31 nm. (b) Fig. 40-17 shows that the width of the Balmer series is 656.3 nm − 364.6 nm ≈ 292 nm. This can be conﬁrmed with a calculation very much like the one shown in Sample Problem 40-6, but with the longest wavelength arising from the 3 → 2 transition, and the series limit obtained from the ∞ → 2 transition. (c) We use Eq. 39-1. For the Lyman series, ∆f = 2.998 × 108 m/s 2.998 × 108 m/s − = 8.2 × 1014 Hz 91.4 × 10−9 m 122 × 10−9 m or 8.2 × 102 THz. For the Balmer series, ∆f = 2.998 × 108 m/s 2.998 × 108 m/s − = 3.65 × 1014 Hz 364.6 × 10−9 m 656.3 × 10−9 m which is equivalent to 365 THz. ...
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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