Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 36, Problem 23P
(a)
To determine
The wavelength for the transition.
(b)
To determine
The wavelength for the transition.
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Students have asked these similar questions
If, in
1
1
= Ry
-
you set ni = 1 and take n2 greater than 1,
you generate what is known as the Lyman
%3D
series.
Find the wavelength of the first mem-
ber of this series.
The value of ħ is
1.05457 × 10¬34 J.s; the Rydberg constant
for hydrogen is 1.09735 × 10’ m¬'; the Bohr
radius is 5.29177 × 10¬1" m; and the ground
state energy for hydrogen is 13.6057 eV.
Answer in units of nm.
Consider the next three members of this se-
ries. The wavelengths of successive members
of the Lyman series approach a common limit
as n2 → ∞.
What is this limit?
Answer in units of nm.
Use the Bohr theory to estimate the wavelength for an n
= 3 to n = 1 transition in molybdenum. The measured
value is 0.063 nm. Why do we not expect perfect
agreement?
The Lyman series comprises a set of spectral lines. All of these lines involve a hydrogen atom whose electron undergoes a change in energy level, either beginning at the n = 1 level (in the case of an absorption line) or ending there (an emission line).
The inverse wavelengths for the Lyman series in hydrogen are given by
1 -
where n = 2, 3, 4, ... and the Rydberg constant R, = 1.097 x 10' m-. (Round your answers to at least one decimal place. Enter your answers in nm.)
%3D
(a) Compute the wavelength for the first line in this series (the line corresponding to n = 2).
nm
(b) Compute the wavelength for the second line in this series (the line corresponding to n = 3).
nm
(c) Compute the wavelength for the third line in this series (the line corresponding to n = 4).
nm
(d) In which part of the electromagnetic spectrum do these three lines reside?
O x-ray region
O ultraviolet region
O infrared region
O gamma ray region
O visible light region
Chapter 36 Solutions
Physics for Scientists and Engineers
Ch. 36 - Prob. 1PCh. 36 - Prob. 2PCh. 36 - Prob. 3PCh. 36 - Prob. 4PCh. 36 - Prob. 5PCh. 36 - Prob. 6PCh. 36 - Prob. 7PCh. 36 - Prob. 8PCh. 36 - Prob. 9PCh. 36 - Prob. 10P
Ch. 36 - Prob. 11PCh. 36 - Prob. 12PCh. 36 - Prob. 13PCh. 36 - Prob. 14PCh. 36 - Prob. 15PCh. 36 - Prob. 16PCh. 36 - Prob. 17PCh. 36 - Prob. 18PCh. 36 - Prob. 19PCh. 36 - Prob. 20PCh. 36 - Prob. 21PCh. 36 - Prob. 22PCh. 36 - Prob. 23PCh. 36 - Prob. 24PCh. 36 - Prob. 25PCh. 36 - Prob. 26PCh. 36 - Prob. 27PCh. 36 - Prob. 28PCh. 36 - Prob. 29PCh. 36 - Prob. 30PCh. 36 - Prob. 31PCh. 36 - Prob. 32PCh. 36 - Prob. 33PCh. 36 - Prob. 34PCh. 36 - Prob. 35PCh. 36 - Prob. 36PCh. 36 - Prob. 37PCh. 36 - Prob. 38PCh. 36 - Prob. 39PCh. 36 - Prob. 40PCh. 36 - Prob. 41PCh. 36 - Prob. 42PCh. 36 - Prob. 43PCh. 36 - Prob. 44PCh. 36 - Prob. 45PCh. 36 - Prob. 46PCh. 36 - Prob. 47PCh. 36 - Prob. 48PCh. 36 - Prob. 49PCh. 36 - Prob. 50PCh. 36 - Prob. 51PCh. 36 - Prob. 52PCh. 36 - Prob. 53PCh. 36 - Prob. 54PCh. 36 - Prob. 55PCh. 36 - Prob. 56PCh. 36 - Prob. 57PCh. 36 - Prob. 58PCh. 36 - Prob. 59PCh. 36 - Prob. 60PCh. 36 - Prob. 61PCh. 36 - Prob. 62PCh. 36 - Prob. 63PCh. 36 - Prob. 64PCh. 36 - Prob. 65PCh. 36 - Prob. 66PCh. 36 - Prob. 67PCh. 36 - Prob. 68PCh. 36 - Prob. 69PCh. 36 - Prob. 70PCh. 36 - Prob. 71PCh. 36 - Prob. 72PCh. 36 - Prob. 73PCh. 36 - Prob. 74PCh. 36 - Prob. 75P
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- Do the Balmer series and the Lyman series overlap? Why? Why not? (Hint: calculate the shortest Balmer line and the longest Lyman line.)arrow_forwardShow that the following 4 lines in the Lyman series can be predicted: 91.127, 97.202, 102.52, and 121.57 nm.arrow_forwardIn the Balmer series, the wavelength of a certain line is measured as 379.1 nm. What transition does this value correspond to?arrow_forward
- Estimate the energy of the characteristic x-ray emitted from a tungsten target when an electron drops from an M shell (n = 3 state) to a vacancy in the K shell (n = 1 state). The atomic number for tungsten is Z = 74.arrow_forward(a) The L→ K transition of an X-ray tube containing a molybdenum (Z = 42) target occurs at a wavelength of 0.0724 nm. Use this information to estimate the screening parameter of the K-shell electrons in molybdenum. [Osmania University]arrow_forward(a) The Lyman series in hydrogen is the transition from energy levels n = 2, 3, 4, ... to the ground state n = 1. The energy levels are given by 13.60 eV En n- (i) What is the second longest wavelength in nm of the Lyman series? (ii) What is the series limit of the Lyman series? [1 eV = 1.602 x 1019 J, h = 6.626 × 10-34 J.s, c = 3 × 10° m.s] %3D Two emission lines have wavelengts A and + A2, respectively, where AA <<2. Show that the angular separation A0 in a grating spectrometer is given aproximately by (b) A0 = V(d/m)-2 where d is the grating constant and m is the order at which the lines are observed.arrow_forward
- The visible lines of the Balmer series were observed first because they are most easily seen. Show that the wavelengths of spectral lines in the Lyman (n = 1) and Paschen (n = 3) series are not in the visible region. Find the wavelengths of the four visible atomic hydrogen lines. Assume the visible wavelength region is λ = 400– 700 nm.arrow_forwardThe K series of the discrete x-ray spectrum of tungsten contains wavelengths of 0.018 5 nm, 0.020 9 nm, and 0.021 5 nm. The K-shell ionization energy is 69.5 keV. (a) Determine the ionization energies of the L, M, and N shells. L shell _keV M shell keV N shell _keVarrow_forwardin a transition from n=~ to n=4, calculate the shortest wavelength photon in the balmer seriesarrow_forward
- It is possible that a muon be captured by a proton to form a muonic atom. A muon is identic to an electron, except when your mass, which is m = 105.7 MeV/c^2. What ia the smallest wave length for a Lyman series for this atom? Give your answer in pm.arrow_forwardThe K series of the discrete spectrum of tungsten contains wavelengths of 0.018 5 nm, 0.020 9 nm, and 0.021 5 nm. The K-shell ionization energy is 69.5 keV. Determine the ionization energies of the L, M, and N shells.arrow_forwardShow that the longest wavelength of the Balmer series and the longest two wavelengths of the Lyman series sat- isfy the Ritz combination principle. For the Lyman series, limit = 91.13 nm.arrow_forward
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