College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 28, Problem 17P
(a)
To determine
The energy of the photon for the transition from
n = 2 n = 5
(b)
To determine
The energy of the photon for the transition from
n = 4 n = 6
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
(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.
The electron in a hydrogen atom with anenergy of -0.544 eV is in a subshell with 18 states. (a) What is theprincipal quantum number, n, for this atom? (b) What is the maximum possible orbital angular momentum this atom can have?(c) Is the number of states in the subshell with the next lowestvalue of / equal to 16, 14, or 12? Explain.
a) Calculate the excitation energies for the 1s → 3p electron transition for the H-atom
and for the He+-ion. Calculate the energy values in joules.
b) What is the ionization energy (in eV) of H-atom in the 4d-state?
Note: En
=
2n²
Eh, 1Eh
= 27.2114eV =
2625.500kJ/mol = 4.35974394-10-18 J
Chapter 28 Solutions
College Physics
Ch. 28.3 - Prob. 28.1QQCh. 28.4 - Prob. 28.2QQCh. 28.5 - Prob. 28.3QQCh. 28 - Prob. 1CQCh. 28 - Prob. 2CQCh. 28 - Prob. 3CQCh. 28 - Prob. 4CQCh. 28 - Prob. 5CQCh. 28 - Prob. 6CQCh. 28 - Prob. 7CQ
Ch. 28 - Prob. 8CQCh. 28 - Prob. 9CQCh. 28 - Prob. 10CQCh. 28 - Prob. 11CQCh. 28 - Prob. 12CQCh. 28 - Prob. 13CQCh. 28 - Prob. 14CQCh. 28 - Prob. 15CQCh. 28 - Prob. 1PCh. 28 - Prob. 2PCh. 28 - Prob. 3PCh. 28 - Prob. 4PCh. 28 - Prob. 5PCh. 28 - Prob. 6PCh. 28 - Prob. 7PCh. 28 - Prob. 8PCh. 28 - Prob. 9PCh. 28 - Prob. 10PCh. 28 - Prob. 11PCh. 28 - Prob. 12PCh. 28 - Prob. 13PCh. 28 - Prob. 14PCh. 28 - Prob. 15PCh. 28 - Prob. 16PCh. 28 - Prob. 17PCh. 28 - Prob. 18PCh. 28 - Prob. 19PCh. 28 - Prob. 20PCh. 28 - Prob. 21PCh. 28 - Prob. 22PCh. 28 - Prob. 23PCh. 28 - Prob. 24PCh. 28 - Prob. 25PCh. 28 - Prob. 26PCh. 28 - Prob. 27PCh. 28 - Prob. 28PCh. 28 - Prob. 29PCh. 28 - Prob. 30PCh. 28 - Prob. 31PCh. 28 - Prob. 32PCh. 28 - Prob. 33PCh. 28 - Prob. 34PCh. 28 - Prob. 35PCh. 28 - Prob. 36PCh. 28 - Prob. 37PCh. 28 - Prob. 38PCh. 28 - Prob. 39PCh. 28 - Prob. 40PCh. 28 - Prob. 41PCh. 28 - Prob. 42PCh. 28 - Prob. 43PCh. 28 - Prob. 44PCh. 28 - Prob. 45PCh. 28 - Prob. 46APCh. 28 - Prob. 47APCh. 28 - Prob. 48APCh. 28 - Prob. 49APCh. 28 - Prob. 50APCh. 28 - Prob. 51APCh. 28 - Prob. 52APCh. 28 - Prob. 53APCh. 28 - Prob. 54AP
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Similar questions
- A photon is emitted during the transition from the n = 7 state to the n = 1 state in the hydrogen atom. Note: Rydberg constant: 2.18x10-18 J Can this photon eject an electron from cesium metal (work function of Cs is 3.42x10-19 J) and why? What would be the speed of that electron once is ejected (mass electron is 9.109x10-31 kg)?arrow_forwardAn alkali metal atom is in the ground state. The orbital angular momentum equals zero and the spin angular momentum is entirely due to the single valence electron. A magnetic field is applied that splits the ground state energy level into two levels, 65 μeV apart.A photon, absorbed by the atom, induces a transition between the two levels. What is the wavelength of the photon? (c = 3.00 × 108 m/s, h = 6.626 × 10-34 J ∙ s, Bohr magneton = μB = 9.27 × 10-24 J/T, 1 eV = 1.60 × 10-19 J) Group of answer choices 19 mm 41 mm 38 mm 25 mm 31 mmarrow_forwardAn electron is in a hydrogen atom with n = 2 and ℓ = 1. (a) Find all the possible angles between the orbital angular momentum vector and the z-axis. (b) Suppose the atom absorbs a photon and rises from the n = 2 and ℓ = 1 state to the n = 3 state. Using conversation of angular momentum, what are the possible values of the final value of ℓ in the n = 3 state?arrow_forward
- An n = 2 shell (L shell) has a 2s state and two 2p states split by the spin-orbit interaction. Careful measurements of the Kα x-ray (n = 2 → Sn = 1) transition reveal only two spectral lines. Explain.arrow_forward(Figure 1) shows a few energy levels of the mercury atom. One valence electron is always in the 6s6s state; the other electron changes states. What transitions are allowed in the emission spectrum? Drag the appropriate items to their respective bins.arrow_forwardAngular momentum and Spin. An electron in an H-atom has orbital angular momentum magnitude and z-component given by L² = 1(1+1)ħ², Lz = m₁h, 1 = 0,1,2,..., n 1 - m₁ = 0, ±1, ±2, ..., ±l 3 S² = s(s+1) h² = =h²₁ 4 Consider an excited electron (n > 1) on an H-atom. The total angular momentum ] = L + Š, whose magnitude and z-component follow a similar dependence to some quantum numbers j and m; as J² = j(j + 1)ħ², Jz = mjħ 1 S₂ = m₂h = ± = h Where j and m; are quantum numbers which assume values that jumps in steps of one such that j is non-negative and −j ≤ m¡ ≤ j. For a given quantum number 1, what are the (two) possible values for j? Clue: we can use the vector sum relation of angular momenta, then consider the z-component only.arrow_forward
- In a hydrogen atom, the small magnetic moment of the proton interacts with the magnetic moment of the electron. This results in a 5.87 meV energy difference between the ms = + 12 and ms = -12 states. What wavelength photon is emitted in a transition between these two states?arrow_forwardWhat is the energy of the photon that, whenabsorbed by a hydrogen atom, could cause anelectronic transition from the n = 3 state tothe n = 5 state?Answer in units of eV. What energy could cause an electronic transition from the n = 5 state to the n = 7state?Answer in units of eV.arrow_forwardAngular momentum and Spin. An electron in an H-atom has orbital angular momentum magnitude and z-component given by L² = 1(1+1)ħ², 1 = 0,1,2,..., n-1 Lz = m₂ħ, m₁ = 0, ±1, ±2,..., ±l 3 S² = s(s+1)h² = h², 4 Consider an excited electron (n > 1) on an H-atom. Sz = msh 1 =+=ħ Show that the minimum angle that the I can have with the z-axis is given by n-1 n L.min = cos Clue: the angle a vector with magnitude V from the z-axis can be computed from cos 0 = V²/Varrow_forward
- 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 regionarrow_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_forwardH-atom. The wave function of one of the electrons in the 2p orbital is given by (ignoring spin) r 2,1,0 (1,0,0)= - 7 exp(-270) c ao 1 |32πα cose Where do is the Bohr radius. In the Bohr model, the radius of the electron orbit is given by m=2 = n²ao = 4ao. The probability that the electron can be found at some radius between r and r + dr is given by 2π P(r) dr = √2 = √ ₁²ª d$ S ² What is the expectation value of the distance of the electron from the nucleus (r)? Clue: expected value is computed by (r) = forP(r) dr then do integration by parts do sin 0 de | Yn.l.m² (r, $,0)|²r² drarrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning