Unformatted text preview: 44. (a) The calculation is shown in Sample Problem 406. 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. 4017 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 406, but with
the longest wavelength arising from the 3 → 2 transition, and the series limit obtained from the
∞ → 2 transition.
(c) We use Eq. 391. 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.
 Fall '08
 SPRUNGER
 Physics

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