Chemistry: The Molecular Nature of Matter and Change
Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781259631757
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 7, Problem 7.73P

(a)

Interpretation Introduction

Interpretation:

The energy level to which a ground state electron in a hydrogen atom jumps to after it absorbs a photon of wavelength 94.91 nm is to be determined.

Concept introduction:

An atom of hydrogen contains one electron. But the spectrum of hydrogen consists of a large number of lines. This is so because a sample of hydrogen contains a very large number of atoms. When energy is supplied to a sample of gaseous atoms of hydrogen, different atoms absorb different amounts of energy. Therefore, the electrons in different atoms jump to different energy levels. Upon losing the energies gained initially, the electrons jump back to lower energy levels and release radiations of different wavelengths.

The equation used to predict the position and wavelength of any line in a given series is called the Rydberg’s equation.

Rydberg’s equation is as follows:

1λ=R(1n121n22)        (1)

Here,

λ  is the wavelength of the line.

n1 and  n2 are positive integers, with n2>n1.

R is the Rydberg’s constant.

The conversion factor to convert wavelength from nm to m is,

1nm=1×109 m

(a)

Expert Solution
Check Mark

Answer to Problem 7.73P

An electron in a hydrogen atom moves to energy level 5 from the ground state upon absorption of a photon of wavelength 94.91 nm.

Explanation of Solution

In the ground state of a hydrogen atom, n=1.

The value of the Rydberg’s constant is 1.096776×107 m1.

Substitute 1 for n1, 1.096776×107 m1 for R and 94.91 nm for λ in equation (1).

1(94.91 nm)(109 m1 nm)=(1.096776×107 m1)(1121n22)19.491×108 m=(1.096776×107 m1)(11n22)0.9606608=(11n22)

Rearrange the above equation and calculate the value of n2 as follows:

1n22=10.96066081n22=0.0393392n22=24.41994n2=5

Conclusion

An electron in a hydrogen atom moves to energy level 5 from the ground state upon absorption of a photon of wavelength 94.91 nm.

(b)

Interpretation Introduction

Interpretation:

The intermediate energy level to which the electron jumps after emission of a photon of wavelength 1281 nm is to be determined.

Concept introduction:

An atom of hydrogen contains one electron. But the spectrum of hydrogen consists of a large number of lines. This is so because a sample of hydrogen contains a very large number of atoms. When energy is supplied to a sample of gaseous atoms of hydrogen, different atoms absorb different amounts of energy. Therefore, the electrons in different atoms jump to different energy levels. Upon losing the energies gained initially, the electrons jump back to lower energy levels and release radiations of different wavelengths.

The equation used to predict the position and wavelength of any line in a given series is called the Rydberg’s equation.

Rydberg’s equation is as follows:

1λ=R(1n121n22)        (1)

Here,

λ  is the wavelength of the line.

n1 and  n2 are positive integers, with n2>n1.

R is the Rydberg’s constant.

The conversion factor to convert wavelength from nm to m is,

1nm=1×109 m

(b)

Expert Solution
Check Mark

Answer to Problem 7.73P

The intermediate energy level to which the electron jumps after emission of a photon of wavelength 1281 nm is n=3.

Explanation of Solution

The emission of a photon leads to the transition of an electron from a higher energy level to a lower energy level. Initially, the electron was in energy level 5. Since in the Rydberg’s equation, n2>n1, therefore n2=5.

The value of the Rydberg’s constant is 1.096776×107 m1.

Substitute 5 for n2, 1.096776×107 m1 for R and 1281 nm for λ in equation (1).

1(1281 nm)(109 m1 nm)=(1.096776×107 m1)(1n12152)11.281×106 m=(1.096776×107 m1)(1n12125)0.07118=(1n12125)

Rearrange the above equation and calculate the value of n1 as follows:

1n12=0.07118+0.040001n12=0.11118n12=8.9944n1=3

Conclusion

The intermediate energy level to which the electron jumps after emission of a photon of wavelength 1281 nm is n=3.

(c)

Interpretation Introduction

Interpretation:

The wavelength of the photon emitted after the electron jumps from n=3 to n=1 is to be determined.

Concept introduction:

An atom of hydrogen contains one electron. But the spectrum of hydrogen consists of a large number of lines. This is so because a sample of hydrogen contains a very large number of atoms. When energy is supplied to a sample of gaseous atoms of hydrogen, different atoms absorb different amounts of energy. Therefore, the electrons in different atoms jump to different energy levels. Upon losing the energies gained initially, the electrons jump back to lower energy levels and release radiations of different wavelengths.

The equation used to predict the position and wavelength of any line in a given series is called the Rydberg’s equation.

Rydberg’s equation is as follows:

1λ=R(1n121n22)        (1)

Here,

λ  is the wavelength of the line.

n1 and  n2 are positive integers, with n2>n1.

R is the Rydberg’s constant.

The conversion factor to convert wavelength from nm to m is,

1nm=1×109 m

(c)

Expert Solution
Check Mark

Answer to Problem 7.73P

The wavelength of the photon emitted after the electron jumps from n=3 to n=1 is 102.6 nm.

Explanation of Solution

The emission of a photon leads to the transition of an electron from a higher energy level to a lower energy level. Initially, the electron was in energy level 3 and later jumps to energy level 1. Since in the Rydberg’s equation, n2>n1, therefore n2=3.

The value of the Rydberg’s constant is 1.096776×107 m1.

Substitute 3 for n2, 1 for n1 and 1.096776×107 m1 for R in equation (1).

1λ=(1.096776×107 m1)(112132)1λ=(1.096776×107 m1)(119)1λ=9.74912×106 m1λ=102.6 nm

Conclusion

The wavelength of the photon emitted after the electron jumps from n=3 to n=1 is 102.6 nm.

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

Chemistry: The Molecular Nature of Matter and Change

Ch. 7.3 - Prob. 7.5AFPCh. 7.3 - Prob. 7.5BFPCh. 7.4 - What are the possible l and m1 values for n = 4? Ch. 7.4 - Prob. 7.6BFPCh. 7.4 - Prob. 7.7AFPCh. 7.4 - Prob. 7.7BFPCh. 7.4 - Prob. 7.8AFPCh. 7.4 - Prob. 7.8BFPCh. 7 - Prob. 7.1PCh. 7 - Consider the following types of electromagnetic...Ch. 7 - Prob. 7.3PCh. 7 - In the 17th century, Isaac Newton proposed that...Ch. 7 - Prob. 7.5PCh. 7 - What new idea about light did Einstein use to...Ch. 7 - An AM station broadcasts rock music at “950 on...Ch. 7 - An FM station broadcasts music at 93.5 MHz...Ch. 7 - Prob. 7.9PCh. 7 - An x-ray has a wavelength of 1.3 Å. Calculate the...Ch. 7 - Prob. 7.11PCh. 7 - Prob. 7.12PCh. 7 - Police often monitor traffic with “K-band” radar...Ch. 7 - Covalent bonds in a molecule absorb radiation in...Ch. 7 - Prob. 7.15PCh. 7 - Prob. 7.16PCh. 7 - How is n1 in the Rydberg equation (Equation 7.4)...Ch. 7 - What key assumption of Bohr’s model would a “Solar...Ch. 7 - Prob. 7.19PCh. 7 - Which of these electron transitions correspond to...Ch. 7 - Why couldn’t the Bohr model predict spectra for...Ch. 7 - Prob. 7.22PCh. 7 - Use the Rydberg equation to find the wavelength...Ch. 7 - Prob. 7.24PCh. 7 - Prob. 7.25PCh. 7 - Prob. 7.26PCh. 7 - Prob. 7.27PCh. 7 - Prob. 7.28PCh. 7 - Prob. 7.29PCh. 7 - Prob. 7.30PCh. 7 - Prob. 7.31PCh. 7 - Prob. 7.32PCh. 7 - In addition to continuous radiation, fluorescent...Ch. 7 - Prob. 7.34PCh. 7 - Prob. 7.35PCh. 7 - Prob. 7.36PCh. 7 - Prob. 7.37PCh. 7 - Prob. 7.38PCh. 7 - A 232-lb fullback runs 40 yd at 19.8 ± 0.1...Ch. 7 - Prob. 7.40PCh. 7 - Prob. 7.41PCh. 7 - Prob. 7.42PCh. 7 - Prob. 7.43PCh. 7 - Prob. 7.44PCh. 7 - What physical meaning is attributed to ψ2? Ch. 7 - What does “electron density in a tiny volume of...Ch. 7 - Prob. 7.47PCh. 7 - Prob. 7.48PCh. 7 - How many orbitals in an atom can have each of the...Ch. 7 - Prob. 7.50PCh. 7 - Give all possible ml values for orbitals that have...Ch. 7 - Prob. 7.52PCh. 7 - Prob. 7.53PCh. 7 - Prob. 7.54PCh. 7 - Prob. 7.55PCh. 7 - Prob. 7.56PCh. 7 - Prob. 7.57PCh. 7 - Prob. 7.58PCh. 7 - Prob. 7.59PCh. 7 - Prob. 7.60PCh. 7 - Prob. 7.61PCh. 7 - The quantum-mechanical treatment of the H atom...Ch. 7 - The photoelectric effect is illustrated in a plot...Ch. 7 - Prob. 7.64PCh. 7 - Prob. 7.65PCh. 7 - Prob. 7.66PCh. 7 - Prob. 7.67PCh. 7 - Prob. 7.68PCh. 7 - Prob. 7.69PCh. 7 - Prob. 7.70PCh. 7 - Prob. 7.71PCh. 7 - Prob. 7.72PCh. 7 - Prob. 7.73PCh. 7 - Prob. 7.74PCh. 7 - Use the relative size of the 3s orbital below to...Ch. 7 - Prob. 7.76PCh. 7 - Prob. 7.77PCh. 7 - Enormous numbers of microwave photons are needed...Ch. 7 - Prob. 7.79PCh. 7 - Prob. 7.80PCh. 7 - Prob. 7.81PCh. 7 - Prob. 7.82PCh. 7 - Prob. 7.83PCh. 7 - Prob. 7.84PCh. 7 - For any microscope, the size of the smallest...Ch. 7 - In fireworks, the heat of the reaction of an...Ch. 7 - Prob. 7.87PCh. 7 - Fish-liver oil is a good source of vitamin A,...Ch. 7 - Many calculators use photocells as their energy...Ch. 7 - Prob. 7.90PCh. 7 - Prob. 7.91PCh. 7 - Prob. 7.92PCh. 7 - The flame tests for sodium and potassium are based...Ch. 7 - Prob. 7.94PCh. 7 - Prob. 7.95PCh. 7 - The discharge of phosphate in detergents to the...
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