General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 7, Problem 7.34QP
Interpretation Introduction

Interpretation:

The value of ni in which the photon emitted that has a wavelength of 434 nm and the involved transition from an energy state of principal quantum number ni to the = 2 state of an electron in the hydrogen atom should be calculated using the concept of Bohr’s theory.

Concept Introduction:

The emission of radiation given by an energized hydrogen atom to the electron falling from a higher-energy orbit to a lower orbit give a quantum of energy in the form of light.  Based on electrostatic interaction and law of motion, Bohr derived the following equation.

En = 2.18 × 1018 J (1n2)

Where, n gets an integer values such as = 1, 2, 3 and so on.  This is the energy of electron in nth orbital.

The electrons are excited thermally when the light is used by an object.  As a result, an emission spectrum comes.  Line spectra consist of light only at specific, discrete wavelengths.  In emission, the electron returns to a lower energy state from nf (the i and f subscripts denote the initial and final energy states).  In most cases, the lower energy state corresponds to the ground state but it may be any energy state which is lower than the initial excited state.  The difference in the energies between the initial and final states is

ΔE = Ef  Ei

This transition results in the photon’s emission with frequency v and energy hv.  The following equation is resulted.

ΔE = hν = 2.18 × 1018 J (1nf21ni2)

When, ni > nf, a photon is emitted.  The term in parentheses is positive, making ΔE negative.  As a result, energy is lost to the surroundings.  When ni < nf, a photon is absorbed.  The term in parentheses is negative, so ΔE is positive.  As a result, energy is absorbed from the surroundings.

The speed, wavelength and frequency of a wave are interrelated by = λν where λ and ν are mentioned in meters (m) and reciprocal seconds (s1).    Hence, rearrange the equation for getting the wavelength is,

ν = cλ

Substitute the frequency formula and rearrange,

ΔE = hν = hcλ = 2.18 × 1018 J (1nf21ni2)1λ = 2.18 × 1018 Jhc (1nf21ni2)

Therefore, this formula is used to find the wavelength of the given photon in the emission line process.  For the absorption line process in which an electron is removed from the nucleus, the sign is changed as,

1λ = 2.18 × 1018 Jhc (1nf21ni2)

To find: Get the value of ni in which the photon emitted that has a wavelength of 434 nm and the involved transition from an energy state of principal quantum number ni to the = 2 state of an electron in the hydrogen atom

Expert Solution & Answer
Check Mark

Answer to Problem 7.34QP

The value of ni in which the photon emitted that has a wavelength of 434 nm and the involved transition from an energy state of principal quantum number ni to the = 2 state of an electron in the hydrogen atom is 5.

Explanation of Solution

An electron in the hydrogen atom makes a transition from an energy state of principal quantum number  ni to the = 2.  The photon emitted has a wavelength of 434 nm.  The formula used to find the value of ni is 

ΔE = hν = 2.18 × 1018 J(1nf21ni2)ΔE = hν = hcλ

Here, Planck’s constant, h is 6.63 × 1034 Js; the speed of light, c is 3.00 × 108 m/s; λ = 434 nm and nf = 2.  To find the value of ni, ΔE should be calculated first. 

ΔE = (6.63 × 1034 Js)(3.00 × 108 m/s)434 × 109 mΔE = 4.58 × 1019 J

Hence, the photon energy change is 4.58 × 1019 J which should be negative because this is an emission process.  Substitute the given values in the formula:

ΔE = 2.18 × 1018 J(1nf21ni2)4.58 × 1019 J = 2.18 × 1018 J(1221ni2)1ni2 = (4.58 × 1019 J2.18 × 1018 J) + 1221ni2 = 0.210 + 0.2501ni2 = 0.040ni = 10.040ni = 5

Therefore, the value of ni in which the photon emitted that has a wavelength of 434 nm and the involved transition from an energy state of principal quantum number ni to the = 2 state of an electron in the hydrogen atom is 5.

Conclusion

The value of ni in which the photon emitted that has a wavelength of 434 nm and the involved transition from an energy state of principal quantum number ni to the = 2 state of an electron in the hydrogen atom should be calculated using the concept of Bohr’s theory.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 7 Solutions

General Chemistry

Ch. 7.5 - Prob. 1RCCh. 7.6 - Prob. 1RCCh. 7.7 - Prob. 1PECh. 7.7 - Prob. 2PECh. 7.7 - Prob. 1RCCh. 7.8 - Prob. 1PECh. 7.8 - Prob. 2PECh. 7.8 - Prob. 3PECh. 7.8 - Prob. 1RCCh. 7.9 - Prob. 1PECh. 7.9 - Prob. 1RCCh. 7 - Prob. 7.1QPCh. 7 - Prob. 7.2QPCh. 7 - Prob. 7.3QPCh. 7 - Prob. 7.4QPCh. 7 - Prob. 7.5QPCh. 7 - Prob. 7.6QPCh. 7 - Prob. 7.7QPCh. 7 - 7.8 (a) What is the frequency of tight having a...Ch. 7 - Prob. 7.9QPCh. 7 - Prob. 7.10QPCh. 7 - Prob. 7.11QPCh. 7 - 7.12 The SI unit of length is the meter, which...Ch. 7 - 7.13 What are photons? What role did Einstein's...Ch. 7 - Prob. 7.14QPCh. 7 - Prob. 7.15QPCh. 7 - Prob. 7.16QPCh. 7 - Prob. 7.17QPCh. 7 - Prob. 7.18QPCh. 7 - Prob. 7.19QPCh. 7 - Prob. 7.20QPCh. 7 - Prob. 7.21QPCh. 7 - Prob. 7.22QPCh. 7 - Prob. 7.23QPCh. 7 - Prob. 7.24QPCh. 7 - Prob. 7.25QPCh. 7 - Prob. 7.26QPCh. 7 - Prob. 7.27QPCh. 7 - Prob. 7.28QPCh. 7 - Prob. 7.29QPCh. 7 - Prob. 7.30QPCh. 7 - Prob. 7.31QPCh. 7 - Prob. 7.32QPCh. 7 - Prob. 7.33QPCh. 7 - Prob. 7.34QPCh. 7 - Prob. 7.35QPCh. 7 - Prob. 7.36QPCh. 7 - Prob. 7.37QPCh. 7 - Prob. 7.38QPCh. 7 - Prob. 7.39QPCh. 7 - Prob. 7.40QPCh. 7 - Prob. 7.41QPCh. 7 - 7.42 What is the de Broglie wavelength (in nm)...Ch. 7 - Prob. 7.43QPCh. 7 - Prob. 7.44QPCh. 7 - Prob. 7.45QPCh. 7 - Prob. 7.46QPCh. 7 - Prob. 7.47QPCh. 7 - Prob. 7.48QPCh. 7 - 7.49 Why is a boundary surface diagram useful in...Ch. 7 - Prob. 7.50QPCh. 7 - Prob. 7.51QPCh. 7 - Prob. 7.52QPCh. 7 - Prob. 7.53QPCh. 7 - Prob. 7.54QPCh. 7 - Prob. 7.55QPCh. 7 - Prob. 7.56QPCh. 7 - Prob. 7.57QPCh. 7 - 7.58 What is the difference between a 2px and a...Ch. 7 - Prob. 7.59QPCh. 7 - Prob. 7.60QPCh. 7 - Prob. 7.61QPCh. 7 - Prob. 7.62QPCh. 7 - Prob. 7.63QPCh. 7 - Prob. 7.64QPCh. 7 - 7.65 Make a chart of all allowable orbitals in the...Ch. 7 - 7.66 Why do the 3s, 3p, and 3d orbitals have the...Ch. 7 - Prob. 7.67QPCh. 7 - Prob. 7.68QPCh. 7 - Prob. 7.69QPCh. 7 - Prob. 7.70QPCh. 7 - Prob. 7.71QPCh. 7 - Prob. 7.72QPCh. 7 - Prob. 7.73QPCh. 7 - Prob. 7.74QPCh. 7 - Prob. 7.75QPCh. 7 - Prob. 7.76QPCh. 7 - Prob. 7.77QPCh. 7 - 7.78 Comment on the correctness of the following...Ch. 7 - Prob. 7.79QPCh. 7 - Prob. 7.80QPCh. 7 - Prob. 7.81QPCh. 7 - Prob. 7.82QPCh. 7 - Prob. 7.83QPCh. 7 - Prob. 7.84QPCh. 7 - Prob. 7.85QPCh. 7 - Prob. 7.86QPCh. 7 - Prob. 7.87QPCh. 7 - Prob. 7.88QPCh. 7 - Prob. 7.89QPCh. 7 - Prob. 7.90QPCh. 7 - Prob. 7.91QPCh. 7 - Prob. 7.92QPCh. 7 - Prob. 7.93QPCh. 7 - Prob. 7.94QPCh. 7 - 7.95 Identify the following individuals and their...Ch. 7 - Prob. 7.96QPCh. 7 - Prob. 7.97QPCh. 7 - Prob. 7.98QPCh. 7 - Prob. 7.99QPCh. 7 - 7.100 A laser is used in treating retina...Ch. 7 - 7.101 A 368-g sample of water absorbs infrared...Ch. 7 - Prob. 7.102QPCh. 7 - Prob. 7.103QPCh. 7 - Prob. 7.104QPCh. 7 - Prob. 7.105QPCh. 7 - Prob. 7.106QPCh. 7 - Prob. 7.107QPCh. 7 - Prob. 7.108QPCh. 7 - Prob. 7.109QPCh. 7 - Prob. 7.110QPCh. 7 - Prob. 7.111QPCh. 7 - 7.112 An atom moving at its root-mean-square speed...Ch. 7 - Prob. 7.113QPCh. 7 - Prob. 7.114QPCh. 7 - Prob. 7.115QPCh. 7 - Prob. 7.116QPCh. 7 - Prob. 7.117SPCh. 7 - Prob. 7.118SPCh. 7 - Prob. 7.119SPCh. 7 - Prob. 7.120SPCh. 7 - 7.121 According to Einstein’s special theory of...Ch. 7 - Prob. 7.122SPCh. 7 - Prob. 7.123SPCh. 7 - Prob. 7.124SPCh. 7 - Prob. 7.125SPCh. 7 - 7.126 The wave function for the 2s orbital in the...Ch. 7 - Prob. 7.127SPCh. 7 - Prob. 7.128SPCh. 7 - Prob. 7.129SP
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Quantum Mechanics - Part 1: Crash Course Physics #43; Author: CrashCourse;https://www.youtube.com/watch?v=7kb1VT0J3DE;License: Standard YouTube License, CC-BY