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
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Chapter 7, Problem 7.82P

(a)

Interpretation Introduction

Interpretation:

The values of ΔE and λ for the photon emitted in the transition of an electron from energy level 3 to level 2 are to be determined.

Concept introduction:

Electromagnetic waves are radiations that are formed by oscillating electric and magnetic fields. The electric and magnetic field components of an electromagnetic wave are perpendicular to each other.

The difference between the energies of two energy levels is the amount of energy in a photon emitted or absorbed when an electron makes a transition. The energy of a photon emitted or absorbed by an electron is,

ΔE=hν

Here,

ΔE is the energy.

h is the Plank’s constant.

ν is the frequency

The equation to relate the frequency and wavelength of radiation is as follows:

ν=cλ

The above relation can be modified as follows:

ΔE=hcλ (1)

(a)

Expert Solution
Check Mark

Answer to Problem 7.82P

The values of ΔE and λ for the photon emitted in the transition of an electron from energy level 3 to level 2 are 7.56×1018 J and 2.63×108 m respectively.

Explanation of Solution

The energies of the electronic transitions E31 and E21 are 4.854×1017 J and 4.098×1017 J respectively.

The formula to calculate the difference in the energies of energy levels 3 and 2 is,

ΔE=E31E21 (2)

Substitute 4.854×1017 J for E31 and 4.098×1017 J for E21 in equation (2).

ΔE=(4.854×1017 J)(4.098×1017 J)=0.756×1017 J=7.56×1018 J

Substitute 7.56×1018 J for ΔE, 3×108 m/s for c and 6.63×1034 Js for h in equation (1) to calculate λ for the photon emitted in the transition.

7.56×1018 J=(6.63×1034 Js)(3×108 m/s)λ

Rearrange the above equation to calculate the value of λ as follows:

λ=(6.63×1034 Js)(3×108 m/s)7.56×1018 J=2.6309×108 m=2.63×108 m

Conclusion

The values of ΔE and λ for the photon emitted in the transition of an electron from energy level 3 to level 2 are 7.56×1018 J and 2.63×108 m respectively.

(b)

Interpretation Introduction

Interpretation:

The values of ΔE and λ for the photon emitted in the transition of an electron from energy level 4 to level 1 are to be determined.

Concept introduction:

Electromagnetic waves are radiations that are formed by oscillating electric and magnetic fields. The electric and magnetic field components of an electromagnetic wave are perpendicular to each other.

The difference between the energies of two energy levels is the amount of energy in a photon emitted or absorbed when an electron makes a transition. The energy of a photon emitted or absorbed by an electron is,

ΔE=hν

Here,

ΔE is the energy.

h is the Plank’s constant.

ν is the frequency

The equation to relate the frequency and wavelength of radiation is as follows:

ν=cλ

The above relation can be modified as follows:

ΔE=hcλ (1)

(b)

Expert Solution
Check Mark

Answer to Problem 7.82P

The values of ΔE and λ for the photon emitted in the transition of an electron from energy level 4 to level 1 are 5.122×1017 J and 3.88×108 m respectively.

Explanation of Solution

The energies of the electronic transitions E42 and E21 are, 1.024×1017 J and 4.098×1017 J respectively.

The formula to calculate the difference in the energies of energy levels 4 and 1 is,

ΔE=E42+E21 (3)

Substitute 1.024×1017 J for E42 and 4.098×1017 J for E21 in equation (3).

ΔE=(1.024×1017 J)+(4.098×1017 J)=5.122×1017 J

Substitute 5.122×1017 J for ΔE, 3×108 m/s for c and 6.63×1034 Js for h in equation (1) to calculate λ for the photon emitted in the transition.

5.122×1017 J=(6.63×1034 Js)(3×108 m/s)λ

Rearrange the above equation to calculate the value of λ as follows:

λ=(6.63×1034 Js)(3×108 m/s)5.122×1017 J=3.883×108 m=3.88×108 m

Conclusion

The values of ΔE and λ for the photon emitted in the transition of an electron from energy level 4 to level 1 are 5.122×1017 J and 3.88×108 m respectively.

(c)

Interpretation Introduction

Interpretation:

The values of ΔE and λ for the photon emitted in the transition of an electron from energy level 5 to level 4 are to be determined.

Concept introduction:

Electromagnetic waves are radiations that are formed by oscillating electric and magnetic fields. The electric and magnetic field components of an electromagnetic wave are perpendicular to each other.

The difference between the energies of two energy levels is the amount of energy in a photon emitted or absorbed when an electron makes a transition. The energy of a photon emitted or absorbed by an electron is,

ΔE=hν

Here,

ΔE is the energy.

h is the Plank’s constant.

ν is the frequency

The equation to relate the frequency and wavelength of radiation is as follows:

ν=cλ

The above relation can be modified as follows:

ΔE=hcλ (1)

(c)

Expert Solution
Check Mark

Answer to Problem 7.82P

The values of ΔE and λ for the photon emitted in the transition of an electron from energy level 5 to level 4 are 1.2×1018 J and 1.66×107 m respectively.

Explanation of Solution

The energies of the electronic transitions E51 and E41 are, 5.242×1017 J and 5.122×1017 J respectively.

The formula to calculate the difference in the energies of energy levels 5 and 4 is,

ΔE=E51E41 (4)

Substitute 5.242×1017 J for E51 and 5.122×1017 J for E41 in equation (4).

ΔE=(5.242×1017 J)(5.122×1017 J)=0.12×1017 J=1.2×1018 J

Substitute 1.2×1018 J for ΔE, 3×108 m/s for c and 6.63×1034 Js for h in equation (1) to calculate λ for the photon emitted in the transition.

1.2×1018 J=(6.63×1034 Js)(3×108 m/s)λ

Rearrange the above equation to calculate the value of λ as follows:

λ=(6.63×1034 Js)(3×108 m/s)1.2×1018 J=16.575×108 m=1.66×107 m

Conclusion

The values of ΔE and λ for the photon emitted in the transition of an electron from energy level 5 to level 4 are 1.2×1018 J and 1.66×107 m respectively.

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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...
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