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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

Ask a homework question - tutors are online