General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 7, Problem 7.29QP

(a)

Interpretation Introduction

Interpretation:

The wavelength of the photon needed to excite an electron from the given energy levels of a hypothetical atom E1 to E4, the energy (in joules) of a photon to excite an electron from E2 to E3 and the wavelength of the photon emitted when an electron drops from the E3 level to the E1 level 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.

(a)

Expert Solution
Check Mark

Answer to Problem 7.29QP

The wavelength of the photon needed to excite an electron from the given energy levels of a hypothetical atom E1 to E4 is 140 nm

Explanation of Solution

To find: Calculate the wavelength of the photon needed to excite an electron from the given energy levels of a hypothetical atom E1 to E4

The given energy levels of a hypothetical atom are given as follows:

E4 = 1.0 × 1019 JE3 = 5.0 × 1019 JE2 = 10 × 1019 JE1 = 15 × 1019 J

The energy difference (ΔE) between E4 and E1 is calculated using the formula:

ΔE = E4  E1ΔE = (1.0 × 1019 J)  (15 × 1019 J)ΔE = 14 × 1019 J

Therefore, the energy difference (ΔE) between E4 and E1 is 14 × 1019 J.  By Bohr’s theory,

ΔE = hνΔE = hcλ      ν = cλλ = hcΔE

Planck’s constant, = 6.63 × 1034 Js and the speed of light, = 3.00 × 108 m/s.  This formula is used to find the wavelength of the photon needed to excite an electron from the given energy levels of a hypothetical atom E1 to E4.  Substitute the given values in the formula:

λ = (6.63 × 1034 Js)(3.00 × 108 m/s)14 × 1019 Jλ = 1.4 × 107 mλ = 1.4 × 102 nm

Therefore, the wavelength of the photon needed to excite an electron from the given energy levels of a hypothetical atom E1 to E4 is 140 nm.

(b)

Interpretation Introduction

Interpretation:

The wavelength of the photon needed to excite an electron from the given energy levels of a hypothetical atom E1 to E4, the energy (in joules) of a photon to excite an electron from E2 to E3 and the wavelength of the photon emitted when an electron drops from the E3 level to the E1 level 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.

(b)

Expert Solution
Check Mark

Answer to Problem 7.29QP

The energy of a photon to excite an electron from E2 to E3 is 5 × 1019 J

Explanation of Solution

To find: Calculate the energy (in joules) a photon must have in order to excite an electron from E2 to E3

The energy difference (ΔE) between E3 and E2 is calculated using the formula:

ΔE = E3  E2

Substitute the given values in the formula:

ΔE = (5.0 × 1019 J)  (10 × 1019 J)ΔE = 5 × 1019 J

Therefore, the energy of a photon to excite an electron from E2 to E3 is 5 × 1019 J.

(c)

Interpretation Introduction

Interpretation:

The wavelength of the photon needed to excite an electron from the given energy levels of a hypothetical atom E1 to E4, the energy (in joules) of a photon to excite an electron from E2 to E3 and the wavelength of the photon emitted when an electron drops from the E3 level to the E1 level 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.

(c)

Expert Solution
Check Mark

Answer to Problem 7.29QP

The wavelength of the photon emitted when an electron drops from the E3 level to the E1  level is 200nm

Explanation of Solution

To find: Calculate the wavelength of the photon emitted when an electron drops from the E3 level to the E1 level

The energy difference (ΔE) between E1 and E3 is calculated using the formula:

ΔE = E E3ΔE = (15 × 1019 J)  (5.0 × 1019 J)ΔE = 10 × 1019 J

Therefore, the energy difference (ΔE) between E1 and E3 is 10 × 1019 J where negative sign is ignored.  By Bohr’s theory,

ΔE = hνΔE = hcλ      ν = cλλ = hcΔE

Planck’s constant, = 6.63 × 1034 Js and the speed of light, = 3.00 × 108 m/s.  This formula is used to find the wavelength of the photon emitted when an electron drops from the E3 level to the E1 level.  Substitute the given values in the formula:

λ = (6.63 × 10-34 Js)(3.00 × 108 m/s)10 × 1019 Jλ = 2.0 × 107 mλ = 2.0 × 102 nm

Therefore, the wavelength of the photon emitted when an electron drops from the E3 level to the E1 level is 200 nm.

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Chapter 7 Solutions

General Chemistry

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