University Physics with Modern Physics, Books a la Carte Edition; Modified MasteringPhysics with Pearson eText -- ValuePack Access Card -- for ... eText -- Valuepack Access Card (14th Edition)
14th Edition
ISBN: 9780134308142
Author: Hugh D. Young, Roger A. Freedman
Publisher: PEARSON
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 39, Problem 39.35E
(a)
To determine
The ratio of number of atoms in
(b)
To determine
The ratio of number of atoms in
(c)
To determine
The ratio of number of atoms in
(d)
To determine
Why emission of
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Hydrogen atoms in the atmosphere of the sun can exist in different energy states. The difference between the lowest energy state (the ground state) and the second to lowest energy state (the first excited state) is about 2.5 eV. The temperature of the sun’s atmosphere is about 5800 K (so kBT = 0.5 eV). What will be the ratio of the number of atoms in the first excited state divided by the number of atoms in the ground state?
One cubic meter of atomic hydrogen at 0°C at atmospheric pressure contains approximately 2.70 × 10 25 atoms. The first excited state of the hydrogen atom has an energy of 10.2 eV above that of the lowest state, called the ground slate. Use the Boltzmann factor to find the number of atoms in the first excited slate (a) at 0 ºC and at (b) (1.00 × 10 4 )°C.
atoms can occupy only certain discrete energy levels. Consider a gas at a temperature of 2 500 K whose atoms can occupy only two energy levels separated by 1.50 eV, where 1 eV (electron volt) is an energy unit equal to 1.60 × 10-19 J. Determine the ratio of the number of atoms in the higher energy level to the number in the lower energy level.
Chapter 39 Solutions
University Physics with Modern Physics, Books a la Carte Edition; Modified MasteringPhysics with Pearson eText -- ValuePack Access Card -- for ... eText -- Valuepack Access Card (14th Edition)
Ch. 39.2 - Prob. 39.2TYUCh. 39.3 - Prob. 39.3TYUCh. 39.4 - Prob. 39.4TYUCh. 39.5 - Prob. 39.5TYUCh. 39.6 - Prob. 39.6TYUCh. 39 - Prob. 39.1DQCh. 39 - Prob. 39.2DQCh. 39 - Prob. 39.3DQCh. 39 - When an electron beam goes through a very small...Ch. 39 - Prob. 39.5DQ
Ch. 39 - Prob. 39.6DQCh. 39 - Prob. 39.7DQCh. 39 - Prob. 39.8DQCh. 39 - Prob. 39.9DQCh. 39 - Prob. 39.10DQCh. 39 - Prob. 39.11DQCh. 39 - Prob. 39.12DQCh. 39 - Prob. 39.13DQCh. 39 - Prob. 39.14DQCh. 39 - Prob. 39.15DQCh. 39 - Prob. 39.16DQCh. 39 - Prob. 39.17DQCh. 39 - Prob. 39.18DQCh. 39 - Prob. 39.19DQCh. 39 - Prob. 39.20DQCh. 39 - Prob. 39.21DQCh. 39 - When you check the air pressure in a tire, a...Ch. 39 - Prob. 39.1ECh. 39 - Prob. 39.2ECh. 39 - Prob. 39.3ECh. 39 - Prob. 39.4ECh. 39 - Prob. 39.5ECh. 39 - Prob. 39.6ECh. 39 - Prob. 39.7ECh. 39 - Prob. 39.8ECh. 39 - Prob. 39.9ECh. 39 - Prob. 39.10ECh. 39 - Prob. 39.11ECh. 39 - Prob. 39.12ECh. 39 - Prob. 39.13ECh. 39 - Prob. 39.14ECh. 39 - Prob. 39.15ECh. 39 - Prob. 39.16ECh. 39 - Prob. 39.17ECh. 39 - Prob. 39.18ECh. 39 - Prob. 39.19ECh. 39 - Prob. 39.20ECh. 39 - Prob. 39.21ECh. 39 - Prob. 39.22ECh. 39 - Prob. 39.23ECh. 39 - Prob. 39.24ECh. 39 - Prob. 39.25ECh. 39 - Prob. 39.26ECh. 39 - Prob. 39.27ECh. 39 - Prob. 39.28ECh. 39 - Prob. 39.29ECh. 39 - Prob. 39.30ECh. 39 - Prob. 39.31ECh. 39 - Prob. 39.32ECh. 39 - Prob. 39.33ECh. 39 - Prob. 39.34ECh. 39 - Prob. 39.35ECh. 39 - Prob. 39.36ECh. 39 - Prob. 39.37ECh. 39 - Prob. 39.38ECh. 39 - Prob. 39.39ECh. 39 - Prob. 39.40ECh. 39 - Prob. 39.41ECh. 39 - Prob. 39.42ECh. 39 - Prob. 39.43ECh. 39 - Prob. 39.44ECh. 39 - Prob. 39.45ECh. 39 - Prob. 39.46ECh. 39 - Prob. 39.47ECh. 39 - Prob. 39.48ECh. 39 - Prob. 39.49ECh. 39 - Prob. 39.50PCh. 39 - Prob. 39.51PCh. 39 - Prob. 39.52PCh. 39 - Prob. 39.53PCh. 39 - Prob. 39.54PCh. 39 - Prob. 39.55PCh. 39 - Prob. 39.56PCh. 39 - Prob. 39.57PCh. 39 - Prob. 39.58PCh. 39 - Prob. 39.59PCh. 39 - An Ideal Blackbody. A large cavity that has a very...Ch. 39 - Prob. 39.61PCh. 39 - Prob. 39.62PCh. 39 - Prob. 39.63PCh. 39 - Prob. 39.64PCh. 39 - Prob. 39.65PCh. 39 - Prob. 39.66PCh. 39 - Prob. 39.67PCh. 39 - Prob. 39.68PCh. 39 - Prob. 39.69PCh. 39 - Prob. 39.70PCh. 39 - Prob. 39.71PCh. 39 - Prob. 39.72PCh. 39 - Prob. 39.73PCh. 39 - Prob. 39.74PCh. 39 - Prob. 39.75PCh. 39 - Prob. 39.76PCh. 39 - Prob. 39.77PCh. 39 - Prob. 39.78PCh. 39 - Prob. 39.79PCh. 39 - Prob. 39.80PCh. 39 - A particle with mass m moves in a potential U(x) =...Ch. 39 - Prob. 39.82PCh. 39 - Prob. 39.83PCh. 39 - DATA In the crystallography lab where you work,...Ch. 39 - Prob. 39.85PCh. 39 - Prob. 39.86CPCh. 39 - Prob. 39.87CPCh. 39 - Prob. 39.88PPCh. 39 - Prob. 39.89PPCh. 39 - Prob. 39.90PPCh. 39 - Prob. 39.91PP
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) The Einstein A coefficient of the 3p² ³P1/2 →3s² ³S₁/2 transition in sodium is 6.14 x 107 s¹. A gas of sodium atoms is excited to the 3p23P1/2 level at time t = 0. What fraction of the atoms are still in the upper level after 20 ns ?arrow_forwardPleasearrow_forwardWhat is the surface temperature of Betelgeuse, a red giant star in the constellation of Orion, which radiates with a peak wavelength of about 970 nm? (b) Rigel, a bluish - white star in Orion, radiates with a peak wavelength of 145 nm. Find the temperature of Rigel’s surface.arrow_forward
- The radius of our Sun is 6.96 x 108 m, and its total power output is 3.85 x 1026 W. Assuming the Sun's surface emits as a black body, calculate its surface temperature.arrow_forwardWhat are the (a) energy, (b) magnitude of the momentum, and (c) wavelength of the photon emitted when a hydrogen atom undergoes a transition from a state with n = 4 to a state with n = 2? (a) Number 2.55 Units eV (b) Number 1.3617 Units kg-m/s or N-s (c) Number 4.865976353 Units This answer has no unitsarrow_forwardThe thermal (black-body) radiation from a star peaks at a wavelength of 300 nm. What is the surface temperature of the star in K?arrow_forward
- 3. Starting from a thermal energy of phonon in the integral form 3π nD U = - ³7 for ( - II) n² dn hwn [exp(ħwn/t) − 1]) 2 where Debye number n = (6N/π)¹/³, find out the high temperature limit of the thermal energy (= 3NT) and heat capacity (= 3N) of phonon. Iarrow_forward2. The five lowest energy levels of a certain atomic gas have the values E₁ = 0, E2= 3.80 eV, E3 = 4.30 eV, E₁ = 7.2 eV, and E5 = 7.5 eV. (a) If the temperature is high enough that all levels are occupied and the gas is illuminated with light of wavelength 2400 nm, what transitions can occur? (b) Which of the transitions you found in part (a) will still occur if the temperature is so low that only the state E₁ is occupied? (c) What wavelength of the incident light would stimulate emission from the state Е₁?arrow_forward5.40x106 atoms are excited to an upper energy level at t = 0 s. At the end of 30.0 ns , 90.0% of these atoms have undergone a quantum jump to the ground state. You may want to review (Pages 1198 - 1200). Part A How many photons have been emitted? ΥΠ ΑΣφ ? photons Submit Previous Answers Request Answer X Incorrect; Try Again; 5 attempts remaining Part B What is the lifetime of the excited state? ην ΑΣφ ? ns Submit Request Answerarrow_forward
- 1arrow_forwardA hydrogen atom ¹H with 200 eV of kinetic energy has a head-on, perfectly elastic collision with a ¹2C atom at rest. Part A Afterward, what is the kinetic energy, in eV, of ¹H? Express your answer in electron volts. KfH = Submit Part B V Kfc = ΑΣΦ Request Answer Afterward, what is the kinetic energy, in eV, of ¹2C? Express your answer in electron volts. ww ΑΣΦ ? ? eV eVarrow_forward3. Given that the internal energy U of blackbody radiation in a cavity of volume V at temperature T is given by U = aVT, where a is independent of both V and T, show by thermodynamic reasoning that the free energy, F, and the entropy, S, of the radiation must have the forms: -jaVT* + Tf(V) F = s = aVT° - {(V) 4 where f(V) is an unknown function of V. Assuming S vanishes at T = 0 (because there is no radiation) or assuming the third law of thermodynamics: as (), as T + 0 av show that the radiation pressure is given by 1 U P = 3 Varrow_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
Physics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
Modern PhysicsPhysicsISBN:9781111794378Author:Raymond A. Serway, Clement J. Moses, Curt A. MoyerPublisher:Cengage Learning
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Physics for Scientists and Engineers with Modern ...
Physics
ISBN:9781337553292
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Modern Physics
Physics
ISBN:9781111794378
Author:Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Publisher:Cengage Learning