(a) The frequency of photon absorbed by the atom.

Question

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Answer 1

Question

Chapter 30, Problem 6TP

To determine

(a)

The frequency of photon absorbed by the atom.

Expert Solution

Answer to Problem 6TP

The frequency of photon absorbed by the atom is 4.59 ×1014 Hz.

Explanation of Solution

Given:

The energy of the level 2 is E2=−3.4 eV.

The energy of the level 3 is E3=−1.5 eV.

Formula used:

The change in energy between initial and final orbit is given by

ΔE=E3−E2

The frequency of photon absorbed by the atom is given by

υ=ΔEh=E3−E2h

Calculation:

The planks constant is h=6.626×10−34 J⋅s.

The frequency of photon absorbed by the atom is calculated as follows:

υ=E3−E2h=( −1.5 eV)−( −3.4 eV)( 6.626× 10 −34 J⋅s)=( 1.9 eV)( 1.6× 10 −19 J 1 eV )( 6.626× 10 −34 J⋅s)=( 3.04× 10 −19 J)( 6.626× 10 −34 J⋅s)

υ=(4.59 × 10 14 s -1)( 1 Hz 1 s -1 )=4.59 ×1014 Hz

Conclusion:

Thus, the frequency of photon absorbed by the atom is 4.59 ×1014 Hz.

To determine

(b)

The frequency of photon emitted by the atom.

Expert Solution

Answer to Problem 6TP

The frequency of photon emitted by the atom is 2.42×1015 Hz.

Explanation of Solution

Given:

The energy of the level 1 is E1=−13.4 eV.

Formula used:

The change in energy between initial and final orbit is given by

ΔE=E2−E1

The frequency of photon emitted by the atom is given by

υ=ΔEh=E2−E1h

Calculation:

The frequency of photon emitted by the atom is calculated as follows:

υ=E2−E1h=( −3.4 eV)−( −13.4 eV)( 6.626× 10 −34 J⋅s)=( 10 eV)( 1.6× 10 −19 J 1 eV )( 6.626× 10 −34 J⋅s)=( 16× 10 −19 J)( 6.626× 10 −34 J⋅s)

υ=(2.42× 10 15 s -1)( 1 Hz 1 s -1 )=2.42×1015 Hz

Conclusion:

Thus, the frequency of photon emitted by the atom is 2.42×1015 Hz.

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Answer 2

Question

Chapter 30, Problem 6TP

To determine

(a)

The frequency of photon absorbed by the atom.

Expert Solution

Answer to Problem 6TP

The frequency of photon absorbed by the atom is 4.59 ×1014 Hz.

Explanation of Solution

Given:

The energy of the level 2 is E2=−3.4 eV.

The energy of the level 3 is E3=−1.5 eV.

Formula used:

The change in energy between initial and final orbit is given by

ΔE=E3−E2

The frequency of photon absorbed by the atom is given by

υ=ΔEh=E3−E2h

Calculation:

The planks constant is h=6.626×10−34 J⋅s.

The frequency of photon absorbed by the atom is calculated as follows:

υ=E3−E2h=( −1.5 eV)−( −3.4 eV)( 6.626× 10 −34 J⋅s)=( 1.9 eV)( 1.6× 10 −19 J 1 eV )( 6.626× 10 −34 J⋅s)=( 3.04× 10 −19 J)( 6.626× 10 −34 J⋅s)

υ=(4.59 × 10 14 s -1)( 1 Hz 1 s -1 )=4.59 ×1014 Hz

Conclusion:

Thus, the frequency of photon absorbed by the atom is 4.59 ×1014 Hz.

To determine

(b)

The frequency of photon emitted by the atom.

Expert Solution

Answer to Problem 6TP

The frequency of photon emitted by the atom is 2.42×1015 Hz.

Explanation of Solution

Given:

The energy of the level 1 is E1=−13.4 eV.

Formula used:

The change in energy between initial and final orbit is given by

ΔE=E2−E1

The frequency of photon emitted by the atom is given by

υ=ΔEh=E2−E1h

Calculation:

The frequency of photon emitted by the atom is calculated as follows:

υ=E2−E1h=( −3.4 eV)−( −13.4 eV)( 6.626× 10 −34 J⋅s)=( 10 eV)( 1.6× 10 −19 J 1 eV )( 6.626× 10 −34 J⋅s)=( 16× 10 −19 J)( 6.626× 10 −34 J⋅s)

υ=(2.42× 10 15 s -1)( 1 Hz 1 s -1 )=2.42×1015 Hz

Conclusion:

Thus, the frequency of photon emitted by the atom is 2.42×1015 Hz.

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Answer 3

Question

Chapter 30, Problem 6TP

To determine

(a)

The frequency of photon absorbed by the atom.

Expert Solution

Answer to Problem 6TP

The frequency of photon absorbed by the atom is 4.59 ×1014 Hz.

Explanation of Solution

Given:

The energy of the level 2 is E2=−3.4 eV.

The energy of the level 3 is E3=−1.5 eV.

Formula used:

The change in energy between initial and final orbit is given by

ΔE=E3−E2

The frequency of photon absorbed by the atom is given by

υ=ΔEh=E3−E2h

Calculation:

The planks constant is h=6.626×10−34 J⋅s.

The frequency of photon absorbed by the atom is calculated as follows:

υ=E3−E2h=( −1.5 eV)−( −3.4 eV)( 6.626× 10 −34 J⋅s)=( 1.9 eV)( 1.6× 10 −19 J 1 eV )( 6.626× 10 −34 J⋅s)=( 3.04× 10 −19 J)( 6.626× 10 −34 J⋅s)

υ=(4.59 × 10 14 s -1)( 1 Hz 1 s -1 )=4.59 ×1014 Hz

Conclusion:

Thus, the frequency of photon absorbed by the atom is 4.59 ×1014 Hz.

To determine

(b)

The frequency of photon emitted by the atom.

Expert Solution

Answer to Problem 6TP

The frequency of photon emitted by the atom is 2.42×1015 Hz.

Explanation of Solution

Given:

The energy of the level 1 is E1=−13.4 eV.

Formula used:

The change in energy between initial and final orbit is given by

ΔE=E2−E1

The frequency of photon emitted by the atom is given by

υ=ΔEh=E2−E1h

Calculation:

The frequency of photon emitted by the atom is calculated as follows:

υ=E2−E1h=( −3.4 eV)−( −13.4 eV)( 6.626× 10 −34 J⋅s)=( 10 eV)( 1.6× 10 −19 J 1 eV )( 6.626× 10 −34 J⋅s)=( 16× 10 −19 J)( 6.626× 10 −34 J⋅s)

υ=(2.42× 10 15 s -1)( 1 Hz 1 s -1 )=2.42×1015 Hz

Conclusion:

Thus, the frequency of photon emitted by the atom is 2.42×1015 Hz.

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Answer 4

Question

Chapter 30, Problem 6TP

To determine

(a)

The frequency of photon absorbed by the atom.

Expert Solution

Answer to Problem 6TP

The frequency of photon absorbed by the atom is 4.59 ×1014 Hz.

Explanation of Solution

Given:

The energy of the level 2 is E2=−3.4 eV.

The energy of the level 3 is E3=−1.5 eV.

Formula used:

The change in energy between initial and final orbit is given by

ΔE=E3−E2

The frequency of photon absorbed by the atom is given by

υ=ΔEh=E3−E2h

Calculation:

The planks constant is h=6.626×10−34 J⋅s.

The frequency of photon absorbed by the atom is calculated as follows:

υ=E3−E2h=( −1.5 eV)−( −3.4 eV)( 6.626× 10 −34 J⋅s)=( 1.9 eV)( 1.6× 10 −19 J 1 eV )( 6.626× 10 −34 J⋅s)=( 3.04× 10 −19 J)( 6.626× 10 −34 J⋅s)

υ=(4.59 × 10 14 s -1)( 1 Hz 1 s -1 )=4.59 ×1014 Hz

Conclusion:

Thus, the frequency of photon absorbed by the atom is 4.59 ×1014 Hz.

To determine

(b)

The frequency of photon emitted by the atom.

Expert Solution

Answer to Problem 6TP

The frequency of photon emitted by the atom is 2.42×1015 Hz.

Explanation of Solution

Given:

The energy of the level 1 is E1=−13.4 eV.

Formula used:

The change in energy between initial and final orbit is given by

ΔE=E2−E1

The frequency of photon emitted by the atom is given by

υ=ΔEh=E2−E1h

Calculation:

The frequency of photon emitted by the atom is calculated as follows:

υ=E2−E1h=( −3.4 eV)−( −13.4 eV)( 6.626× 10 −34 J⋅s)=( 10 eV)( 1.6× 10 −19 J 1 eV )( 6.626× 10 −34 J⋅s)=( 16× 10 −19 J)( 6.626× 10 −34 J⋅s)

υ=(2.42× 10 15 s -1)( 1 Hz 1 s -1 )=2.42×1015 Hz

Conclusion:

Thus, the frequency of photon emitted by the atom is 2.42×1015 Hz.

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Answer 5

Chapter 30, Problem 6TP

To determine

(a)

The frequency of photon absorbed by the atom.

Expert Solution

Answer to Problem 6TP

The frequency of photon absorbed by the atom is 4.59 ×1014 Hz.

Explanation of Solution

Given:

The energy of the level 2 is E2=−3.4 eV.

The energy of the level 3 is E3=−1.5 eV.

Formula used:

The change in energy between initial and final orbit is given by

ΔE=E3−E2

The frequency of photon absorbed by the atom is given by

υ=ΔEh=E3−E2h

Calculation:

The planks constant is h=6.626×10−34 J⋅s.

The frequency of photon absorbed by the atom is calculated as follows:

υ=E3−E2h=( −1.5 eV)−( −3.4 eV)( 6.626× 10 −34 J⋅s)=( 1.9 eV)( 1.6× 10 −19 J 1 eV )( 6.626× 10 −34 J⋅s)=( 3.04× 10 −19 J)( 6.626× 10 −34 J⋅s)

υ=(4.59 × 10 14 s -1)( 1 Hz 1 s -1 )=4.59 ×1014 Hz

Conclusion:

Thus, the frequency of photon absorbed by the atom is 4.59 ×1014 Hz.

To determine

(b)

The frequency of photon emitted by the atom.

Expert Solution

Answer to Problem 6TP

The frequency of photon emitted by the atom is 2.42×1015 Hz.

Explanation of Solution

Given:

The energy of the level 1 is E1=−13.4 eV.

Formula used:

The change in energy between initial and final orbit is given by

ΔE=E2−E1

The frequency of photon emitted by the atom is given by

υ=ΔEh=E2−E1h

Calculation:

The frequency of photon emitted by the atom is calculated as follows:

υ=E2−E1h=( −3.4 eV)−( −13.4 eV)( 6.626× 10 −34 J⋅s)=( 10 eV)( 1.6× 10 −19 J 1 eV )( 6.626× 10 −34 J⋅s)=( 16× 10 −19 J)( 6.626× 10 −34 J⋅s)

υ=(2.42× 10 15 s -1)( 1 Hz 1 s -1 )=2.42×1015 Hz

Conclusion:

Thus, the frequency of photon emitted by the atom is 2.42×1015 Hz.

Answer 6

Chapter 30, Problem 6TP

To determine

(a)

The frequency of photon absorbed by the atom.

Expert Solution

Answer to Problem 6TP

The frequency of photon absorbed by the atom is 4.59 ×1014 Hz.

Explanation of Solution

Given:

The energy of the level 2 is E2=−3.4 eV.

The energy of the level 3 is E3=−1.5 eV.

Formula used:

The change in energy between initial and final orbit is given by

ΔE=E3−E2

The frequency of photon absorbed by the atom is given by

υ=ΔEh=E3−E2h

Calculation:

The planks constant is h=6.626×10−34 J⋅s.

The frequency of photon absorbed by the atom is calculated as follows:

υ=E3−E2h=( −1.5 eV)−( −3.4 eV)( 6.626× 10 −34 J⋅s)=( 1.9 eV)( 1.6× 10 −19 J 1 eV )( 6.626× 10 −34 J⋅s)=( 3.04× 10 −19 J)( 6.626× 10 −34 J⋅s)

υ=(4.59 × 10 14 s -1)( 1 Hz 1 s -1 )=4.59 ×1014 Hz

Conclusion:

Thus, the frequency of photon absorbed by the atom is 4.59 ×1014 Hz.

Answer 7

Chapter 30, Problem 6TP

To determine

(a)

The frequency of photon absorbed by the atom.

Answer 8

Expert Solution

Answer to Problem 6TP

The frequency of photon absorbed by the atom is 4.59 ×1014 Hz.

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

The energy of the level 2 is E2=−3.4 eV.

The energy of the level 3 is E3=−1.5 eV.

Formula used:

The change in energy between initial and final orbit is given by

ΔE=E3−E2

The frequency of photon absorbed by the atom is given by

υ=ΔEh=E3−E2h

Calculation:

The planks constant is h=6.626×10−34 J⋅s.

The frequency of photon absorbed by the atom is calculated as follows:

υ=E3−E2h=( −1.5 eV)−( −3.4 eV)( 6.626× 10 −34 J⋅s)=( 1.9 eV)( 1.6× 10 −19 J 1 eV )( 6.626× 10 −34 J⋅s)=( 3.04× 10 −19 J)( 6.626× 10 −34 J⋅s)

υ=(4.59 × 10 14 s -1)( 1 Hz 1 s -1 )=4.59 ×1014 Hz

Conclusion:

Thus, the frequency of photon absorbed by the atom is 4.59 ×1014 Hz.

Answer 13

To determine

(b)

The frequency of photon emitted by the atom.

Expert Solution

Answer to Problem 6TP

The frequency of photon emitted by the atom is 2.42×1015 Hz.

Explanation of Solution

Given:

The energy of the level 1 is E1=−13.4 eV.

Formula used:

The change in energy between initial and final orbit is given by

ΔE=E2−E1

The frequency of photon emitted by the atom is given by

υ=ΔEh=E2−E1h

Calculation:

The frequency of photon emitted by the atom is calculated as follows:

υ=E2−E1h=( −3.4 eV)−( −13.4 eV)( 6.626× 10 −34 J⋅s)=( 10 eV)( 1.6× 10 −19 J 1 eV )( 6.626× 10 −34 J⋅s)=( 16× 10 −19 J)( 6.626× 10 −34 J⋅s)

υ=(2.42× 10 15 s -1)( 1 Hz 1 s -1 )=2.42×1015 Hz

Conclusion:

Thus, the frequency of photon emitted by the atom is 2.42×1015 Hz.

Answer 14

To determine

(b)

The frequency of photon emitted by the atom.

Answer 15

Expert Solution

Answer to Problem 6TP

The frequency of photon emitted by the atom is 2.42×1015 Hz.

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given:

The energy of the level 1 is E1=−13.4 eV.

Formula used:

The change in energy between initial and final orbit is given by

ΔE=E2−E1

The frequency of photon emitted by the atom is given by

υ=ΔEh=E2−E1h

Calculation:

The frequency of photon emitted by the atom is calculated as follows:

υ=E2−E1h=( −3.4 eV)−( −13.4 eV)( 6.626× 10 −34 J⋅s)=( 10 eV)( 1.6× 10 −19 J 1 eV )( 6.626× 10 −34 J⋅s)=( 16× 10 −19 J)( 6.626× 10 −34 J⋅s)

υ=(2.42× 10 15 s -1)( 1 Hz 1 s -1 )=2.42×1015 Hz

Conclusion:

Thus, the frequency of photon emitted by the atom is 2.42×1015 Hz.

Answer 20

Ch. 30 - Name three different types of evidence for the...Ch. 30 - Explain why patterns observed in the periodic...Ch. 30 - If atoms exist, why can't we see them with visible...Ch. 30 - What two pieces of evidence allowed the first...Ch. 30 - How do the allowed orbits for electrons in atoms...Ch. 30 - How do the allowed orbits for electrons in atoms...Ch. 30 - Explain how Bohr's rule for the quantization of...Ch. 30 - What is a hydrogen-like atom, and how are the...Ch. 30 - Explain why characteristic x rays are the most...Ch. 30 - Why does the energy of characteristic x rays...

(a) The frequency of photon absorbed by the atom.

(a) The frequency of photon absorbed by the atom.

Answer to Problem 6TP

Explanation of Solution

(b) The frequency of photon emitted by the atom.

Answer to Problem 6TP

Explanation of Solution

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Chapter 30 Solutions

(a)

The frequency of photon absorbed by the atom.

(b)

The frequency of photon emitted by the atom.