CHEMISTRY >CUSTOM<
CHEMISTRY >CUSTOM<
8th Edition
ISBN: 9781309097182
Author: SILBERBERG
Publisher: MCG/CREATE
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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 >CUSTOM<

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