The maximum wavelength of radiation .

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 31, Problem 67P

(a)

To determine

The maximum wavelength of radiation.

(a)

Expert Solution

Answer to Problem 67P

The maximum wavelength of radiation is 588 nm .

Explanation of Solution

Given:

The energy at ground state is, E0=0 .

The first excited of atoms is E1=2.11 eV .

The second excited of atoms is E2=3.20 eV .

The third excited of atoms is E3=4.35 eV .

Formula used:

The maximum wavelength of radiation is given as,

λ=h×cΔE

Here, h is the Planck’s constant and its value is 6.26×10−34 J⋅s , c is the speed of light and its value 3×108 m/s and ΔE is the difference in the energy state.

Calculation:

The maximum wavelength of radiation from state 2 to ground state can be calculated as,

λ2−0=h×cE2−E0=6.626× 10 −34 J⋅s×3× 108 m/s( 3.20 eV×( 1.6× 10 −19 J 1 eV )−0)=388×10−9 m( 10 9 nm 1 m)=388 nm

The wavelength of radiation from state 2 to state 1 can be calculated as,

λ2−1=h×cE2−E1=6.626× 10 −34 J⋅s×3× 108 m/s( 3.20 eV×( 1.6× 10 −19 J 1 eV )−2.11 eV×( 1.6× 10 −19 J 1 eV ))=1140×10−9 m( 10 9 nm 1 m)=1140 nm

The maximum wavelength of radiation from state 1 to ground state can be calculated as,

λ1−0=h×cE1−E0=6.626× 10 −34 J⋅s×3× 108 m/s( 2.11 eV×( 1.6× 10 −19 J 1 eV )−0)=588×10−9 m( 10 9 nm 1 m)=588 nm

Conclusion:

Therefore, the maximum wavelength of radiation for resonance absorption is 588 nm .

(b)

To determine

The wavelength at excitation state of 4.35 eV and the wavelength of radiation at excited states.

(b)

Expert Solution

Answer to Problem 67P

The wavelength at excitation state of 4.35 eV is 285 nm , the wavelengths of resonance fluorescence at excited state are 554 nm and 558 nm

Explanation of Solution

Calculation:

The wavelength of radiation from excited state 3 to state 1 can be calculated as,

λ3−1=h×cE3−E1=6.626× 10 −34 J⋅s×3× 108 m/s( 4.35 eV×( 1.6× 10 −19 J 1 eV )−2.11 eV×( 1.6× 10 −19 J 1 eV ))=554×10−9 m( 10 9 nm 1 m)=554 nm

The wavelength of radiation from excited state 1 to ground state can be calculated as,

λ1−0=h×cE1−E0=6.626× 10 −34 J⋅s×3× 108 m/s( 2.11 eV×( 1.6× 10 −19 J 1 eV )−0)=558×10−9 m( 10 9 nm 1 m)=558 nm

Conclusion:

Therefore, the wavelength at excitation state of 4.35 eV is 285 nm , the wavelengths of resonance fluorescence at excited state are 554 nm and 558 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!

Students have asked these similar questions

Four capacitors are connected as shown in the figure below. (Let C = 12.0 μF.) a C 3.00 με Hh. 6.00 με 20.0 με HE (a) Find the equivalent capacitance between points a and b. 5.92 HF (b) Calculate the charge on each capacitor, taking AV ab = 16.0 V. 20.0 uF capacitor 94.7 6.00 uF capacitor 67.6 32.14 3.00 µF capacitor capacitor C ☑ με με The 3 µF and 12.0 uF capacitors are in series and that combination is in parallel with the 6 μF capacitor. What quantity is the same for capacitors in parallel? μC 32.14 ☑ You are correct that the charge on this capacitor will be the same as the charge on the 3 μF capacitor. μC

In the pivot assignment, we observed waves moving on a string stretched by hanging weights. We noticed that certain frequencies produced standing waves. One such situation is shown below: 0 ст Direct Measurement ©2015 Peter Bohacek I. 20 0 cm 10 20 30 40 50 60 70 80 90 100 Which Harmonic is this? Do NOT include units! What is the wavelength of this wave in cm with only no decimal places? If the speed of this wave is 2500 cm/s, what is the frequency of this harmonic (in Hz, with NO decimal places)?

Four capacitors are connected as shown in the figure below. (Let C = 12.0 µF.) A circuit consists of four capacitors. It begins at point a before the wire splits in two directions. On the upper split, there is a capacitor C followed by a 3.00 µF capacitor. On the lower split, there is a 6.00 µF capacitor. The two splits reconnect and are followed by a 20.0 µF capacitor, which is then followed by point b. (a) Find the equivalent capacitance between points a and b. µF(b) Calculate the charge on each capacitor, taking ΔVab = 16.0 V. 20.0 µF capacitor µC 6.00 µF capacitor µC 3.00 µF capacitor µC capacitor C µC

Answer 2

Question

Chapter 31, Problem 67P

(a)

To determine

The maximum wavelength of radiation.

(a)

Expert Solution

Answer to Problem 67P

The maximum wavelength of radiation is 588 nm .

Explanation of Solution

Given:

The energy at ground state is, E0=0 .

The first excited of atoms is E1=2.11 eV .

The second excited of atoms is E2=3.20 eV .

The third excited of atoms is E3=4.35 eV .

Formula used:

The maximum wavelength of radiation is given as,

λ=h×cΔE

Here, h is the Planck’s constant and its value is 6.26×10−34 J⋅s , c is the speed of light and its value 3×108 m/s and ΔE is the difference in the energy state.

Calculation:

The maximum wavelength of radiation from state 2 to ground state can be calculated as,

λ2−0=h×cE2−E0=6.626× 10 −34 J⋅s×3× 108 m/s( 3.20 eV×( 1.6× 10 −19 J 1 eV )−0)=388×10−9 m( 10 9 nm 1 m)=388 nm

The wavelength of radiation from state 2 to state 1 can be calculated as,

λ2−1=h×cE2−E1=6.626× 10 −34 J⋅s×3× 108 m/s( 3.20 eV×( 1.6× 10 −19 J 1 eV )−2.11 eV×( 1.6× 10 −19 J 1 eV ))=1140×10−9 m( 10 9 nm 1 m)=1140 nm

The maximum wavelength of radiation from state 1 to ground state can be calculated as,

λ1−0=h×cE1−E0=6.626× 10 −34 J⋅s×3× 108 m/s( 2.11 eV×( 1.6× 10 −19 J 1 eV )−0)=588×10−9 m( 10 9 nm 1 m)=588 nm

Conclusion:

Therefore, the maximum wavelength of radiation for resonance absorption is 588 nm .

(b)

To determine

The wavelength at excitation state of 4.35 eV and the wavelength of radiation at excited states.

(b)

Expert Solution

Answer to Problem 67P

The wavelength at excitation state of 4.35 eV is 285 nm , the wavelengths of resonance fluorescence at excited state are 554 nm and 558 nm

Explanation of Solution

Calculation:

The wavelength of radiation from excited state 3 to state 1 can be calculated as,

λ3−1=h×cE3−E1=6.626× 10 −34 J⋅s×3× 108 m/s( 4.35 eV×( 1.6× 10 −19 J 1 eV )−2.11 eV×( 1.6× 10 −19 J 1 eV ))=554×10−9 m( 10 9 nm 1 m)=554 nm

The wavelength of radiation from excited state 1 to ground state can be calculated as,

λ1−0=h×cE1−E0=6.626× 10 −34 J⋅s×3× 108 m/s( 2.11 eV×( 1.6× 10 −19 J 1 eV )−0)=558×10−9 m( 10 9 nm 1 m)=558 nm

Conclusion:

Therefore, the wavelength at excitation state of 4.35 eV is 285 nm , the wavelengths of resonance fluorescence at excited state are 554 nm and 558 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!

Answer 3

Question

Chapter 31, Problem 67P

(a)

To determine

The maximum wavelength of radiation.

(a)

Expert Solution

Answer to Problem 67P

The maximum wavelength of radiation is 588 nm .

Explanation of Solution

Given:

The energy at ground state is, E0=0 .

The first excited of atoms is E1=2.11 eV .

The second excited of atoms is E2=3.20 eV .

The third excited of atoms is E3=4.35 eV .

Formula used:

The maximum wavelength of radiation is given as,

λ=h×cΔE

Here, h is the Planck’s constant and its value is 6.26×10−34 J⋅s , c is the speed of light and its value 3×108 m/s and ΔE is the difference in the energy state.

Calculation:

The maximum wavelength of radiation from state 2 to ground state can be calculated as,

λ2−0=h×cE2−E0=6.626× 10 −34 J⋅s×3× 108 m/s( 3.20 eV×( 1.6× 10 −19 J 1 eV )−0)=388×10−9 m( 10 9 nm 1 m)=388 nm

The wavelength of radiation from state 2 to state 1 can be calculated as,

λ2−1=h×cE2−E1=6.626× 10 −34 J⋅s×3× 108 m/s( 3.20 eV×( 1.6× 10 −19 J 1 eV )−2.11 eV×( 1.6× 10 −19 J 1 eV ))=1140×10−9 m( 10 9 nm 1 m)=1140 nm

The maximum wavelength of radiation from state 1 to ground state can be calculated as,

λ1−0=h×cE1−E0=6.626× 10 −34 J⋅s×3× 108 m/s( 2.11 eV×( 1.6× 10 −19 J 1 eV )−0)=588×10−9 m( 10 9 nm 1 m)=588 nm

Conclusion:

Therefore, the maximum wavelength of radiation for resonance absorption is 588 nm .

(b)

To determine

The wavelength at excitation state of 4.35 eV and the wavelength of radiation at excited states.

(b)

Expert Solution

Answer to Problem 67P

The wavelength at excitation state of 4.35 eV is 285 nm , the wavelengths of resonance fluorescence at excited state are 554 nm and 558 nm

Explanation of Solution

Calculation:

The wavelength of radiation from excited state 3 to state 1 can be calculated as,

λ3−1=h×cE3−E1=6.626× 10 −34 J⋅s×3× 108 m/s( 4.35 eV×( 1.6× 10 −19 J 1 eV )−2.11 eV×( 1.6× 10 −19 J 1 eV ))=554×10−9 m( 10 9 nm 1 m)=554 nm

The wavelength of radiation from excited state 1 to ground state can be calculated as,

λ1−0=h×cE1−E0=6.626× 10 −34 J⋅s×3× 108 m/s( 2.11 eV×( 1.6× 10 −19 J 1 eV )−0)=558×10−9 m( 10 9 nm 1 m)=558 nm

Conclusion:

Therefore, the wavelength at excitation state of 4.35 eV is 285 nm , the wavelengths of resonance fluorescence at excited state are 554 nm and 558 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!

Answer 4

Question

Chapter 31, Problem 67P

(a)

To determine

The maximum wavelength of radiation.

(a)

Expert Solution

Answer to Problem 67P

The maximum wavelength of radiation is 588 nm .

Explanation of Solution

Given:

The energy at ground state is, E0=0 .

The first excited of atoms is E1=2.11 eV .

The second excited of atoms is E2=3.20 eV .

The third excited of atoms is E3=4.35 eV .

Formula used:

The maximum wavelength of radiation is given as,

λ=h×cΔE

Here, h is the Planck’s constant and its value is 6.26×10−34 J⋅s , c is the speed of light and its value 3×108 m/s and ΔE is the difference in the energy state.

Calculation:

The maximum wavelength of radiation from state 2 to ground state can be calculated as,

λ2−0=h×cE2−E0=6.626× 10 −34 J⋅s×3× 108 m/s( 3.20 eV×( 1.6× 10 −19 J 1 eV )−0)=388×10−9 m( 10 9 nm 1 m)=388 nm

The wavelength of radiation from state 2 to state 1 can be calculated as,

λ2−1=h×cE2−E1=6.626× 10 −34 J⋅s×3× 108 m/s( 3.20 eV×( 1.6× 10 −19 J 1 eV )−2.11 eV×( 1.6× 10 −19 J 1 eV ))=1140×10−9 m( 10 9 nm 1 m)=1140 nm

The maximum wavelength of radiation from state 1 to ground state can be calculated as,

λ1−0=h×cE1−E0=6.626× 10 −34 J⋅s×3× 108 m/s( 2.11 eV×( 1.6× 10 −19 J 1 eV )−0)=588×10−9 m( 10 9 nm 1 m)=588 nm

Conclusion:

Therefore, the maximum wavelength of radiation for resonance absorption is 588 nm .

(b)

To determine

The wavelength at excitation state of 4.35 eV and the wavelength of radiation at excited states.

(b)

Expert Solution

Answer to Problem 67P

The wavelength at excitation state of 4.35 eV is 285 nm , the wavelengths of resonance fluorescence at excited state are 554 nm and 558 nm

Explanation of Solution

Calculation:

The wavelength of radiation from excited state 3 to state 1 can be calculated as,

λ3−1=h×cE3−E1=6.626× 10 −34 J⋅s×3× 108 m/s( 4.35 eV×( 1.6× 10 −19 J 1 eV )−2.11 eV×( 1.6× 10 −19 J 1 eV ))=554×10−9 m( 10 9 nm 1 m)=554 nm

The wavelength of radiation from excited state 1 to ground state can be calculated as,

λ1−0=h×cE1−E0=6.626× 10 −34 J⋅s×3× 108 m/s( 2.11 eV×( 1.6× 10 −19 J 1 eV )−0)=558×10−9 m( 10 9 nm 1 m)=558 nm

Conclusion:

Therefore, the wavelength at excitation state of 4.35 eV is 285 nm , the wavelengths of resonance fluorescence at excited state are 554 nm and 558 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!

Answer 5

Chapter 31, Problem 67P

(a)

To determine

The maximum wavelength of radiation.

(a)

Expert Solution

Answer to Problem 67P

The maximum wavelength of radiation is 588 nm .

Explanation of Solution

Given:

The energy at ground state is, E0=0 .

The first excited of atoms is E1=2.11 eV .

The second excited of atoms is E2=3.20 eV .

The third excited of atoms is E3=4.35 eV .

Formula used:

The maximum wavelength of radiation is given as,

λ=h×cΔE

Here, h is the Planck’s constant and its value is 6.26×10−34 J⋅s , c is the speed of light and its value 3×108 m/s and ΔE is the difference in the energy state.

Calculation:

The maximum wavelength of radiation from state 2 to ground state can be calculated as,

λ2−0=h×cE2−E0=6.626× 10 −34 J⋅s×3× 108 m/s( 3.20 eV×( 1.6× 10 −19 J 1 eV )−0)=388×10−9 m( 10 9 nm 1 m)=388 nm

The wavelength of radiation from state 2 to state 1 can be calculated as,

λ2−1=h×cE2−E1=6.626× 10 −34 J⋅s×3× 108 m/s( 3.20 eV×( 1.6× 10 −19 J 1 eV )−2.11 eV×( 1.6× 10 −19 J 1 eV ))=1140×10−9 m( 10 9 nm 1 m)=1140 nm

The maximum wavelength of radiation from state 1 to ground state can be calculated as,

λ1−0=h×cE1−E0=6.626× 10 −34 J⋅s×3× 108 m/s( 2.11 eV×( 1.6× 10 −19 J 1 eV )−0)=588×10−9 m( 10 9 nm 1 m)=588 nm

Conclusion:

Therefore, the maximum wavelength of radiation for resonance absorption is 588 nm .

(b)

To determine

The wavelength at excitation state of 4.35 eV and the wavelength of radiation at excited states.

(b)

Expert Solution

Answer to Problem 67P

The wavelength at excitation state of 4.35 eV is 285 nm , the wavelengths of resonance fluorescence at excited state are 554 nm and 558 nm

Explanation of Solution

Calculation:

The wavelength of radiation from excited state 3 to state 1 can be calculated as,

λ3−1=h×cE3−E1=6.626× 10 −34 J⋅s×3× 108 m/s( 4.35 eV×( 1.6× 10 −19 J 1 eV )−2.11 eV×( 1.6× 10 −19 J 1 eV ))=554×10−9 m( 10 9 nm 1 m)=554 nm

The wavelength of radiation from excited state 1 to ground state can be calculated as,

λ1−0=h×cE1−E0=6.626× 10 −34 J⋅s×3× 108 m/s( 2.11 eV×( 1.6× 10 −19 J 1 eV )−0)=558×10−9 m( 10 9 nm 1 m)=558 nm

Conclusion:

Therefore, the wavelength at excitation state of 4.35 eV is 285 nm , the wavelengths of resonance fluorescence at excited state are 554 nm and 558 nm

Answer 6

Chapter 31, Problem 67P

(a)

To determine

The maximum wavelength of radiation.

(a)

Expert Solution

Answer to Problem 67P

The maximum wavelength of radiation is 588 nm .

Explanation of Solution

Given:

The energy at ground state is, E0=0 .

The first excited of atoms is E1=2.11 eV .

The second excited of atoms is E2=3.20 eV .

The third excited of atoms is E3=4.35 eV .

Formula used:

The maximum wavelength of radiation is given as,

λ=h×cΔE

Here, h is the Planck’s constant and its value is 6.26×10−34 J⋅s , c is the speed of light and its value 3×108 m/s and ΔE is the difference in the energy state.

Calculation:

The maximum wavelength of radiation from state 2 to ground state can be calculated as,

λ2−0=h×cE2−E0=6.626× 10 −34 J⋅s×3× 108 m/s( 3.20 eV×( 1.6× 10 −19 J 1 eV )−0)=388×10−9 m( 10 9 nm 1 m)=388 nm

The wavelength of radiation from state 2 to state 1 can be calculated as,

λ2−1=h×cE2−E1=6.626× 10 −34 J⋅s×3× 108 m/s( 3.20 eV×( 1.6× 10 −19 J 1 eV )−2.11 eV×( 1.6× 10 −19 J 1 eV ))=1140×10−9 m( 10 9 nm 1 m)=1140 nm

The maximum wavelength of radiation from state 1 to ground state can be calculated as,

λ1−0=h×cE1−E0=6.626× 10 −34 J⋅s×3× 108 m/s( 2.11 eV×( 1.6× 10 −19 J 1 eV )−0)=588×10−9 m( 10 9 nm 1 m)=588 nm

Conclusion:

Therefore, the maximum wavelength of radiation for resonance absorption is 588 nm .

Answer 7

Chapter 31, Problem 67P

(a)

To determine

The maximum wavelength of radiation.

Answer 8

(a)

Expert Solution

Answer to Problem 67P

The maximum wavelength of radiation is 588 nm .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

The energy at ground state is, E0=0 .

The first excited of atoms is E1=2.11 eV .

The second excited of atoms is E2=3.20 eV .

The third excited of atoms is E3=4.35 eV .

Formula used:

The maximum wavelength of radiation is given as,

λ=h×cΔE

Here, h is the Planck’s constant and its value is 6.26×10−34 J⋅s , c is the speed of light and its value 3×108 m/s and ΔE is the difference in the energy state.

Calculation:

The maximum wavelength of radiation from state 2 to ground state can be calculated as,

λ2−0=h×cE2−E0=6.626× 10 −34 J⋅s×3× 108 m/s( 3.20 eV×( 1.6× 10 −19 J 1 eV )−0)=388×10−9 m( 10 9 nm 1 m)=388 nm

The wavelength of radiation from state 2 to state 1 can be calculated as,

λ2−1=h×cE2−E1=6.626× 10 −34 J⋅s×3× 108 m/s( 3.20 eV×( 1.6× 10 −19 J 1 eV )−2.11 eV×( 1.6× 10 −19 J 1 eV ))=1140×10−9 m( 10 9 nm 1 m)=1140 nm

The maximum wavelength of radiation from state 1 to ground state can be calculated as,

λ1−0=h×cE1−E0=6.626× 10 −34 J⋅s×3× 108 m/s( 2.11 eV×( 1.6× 10 −19 J 1 eV )−0)=588×10−9 m( 10 9 nm 1 m)=588 nm

Conclusion:

Therefore, the maximum wavelength of radiation for resonance absorption is 588 nm .

Answer 13

(b)

To determine

The wavelength at excitation state of 4.35 eV and the wavelength of radiation at excited states.

(b)

Expert Solution

Answer to Problem 67P

The wavelength at excitation state of 4.35 eV is 285 nm , the wavelengths of resonance fluorescence at excited state are 554 nm and 558 nm

Explanation of Solution

Calculation:

The wavelength of radiation from excited state 3 to state 1 can be calculated as,

λ3−1=h×cE3−E1=6.626× 10 −34 J⋅s×3× 108 m/s( 4.35 eV×( 1.6× 10 −19 J 1 eV )−2.11 eV×( 1.6× 10 −19 J 1 eV ))=554×10−9 m( 10 9 nm 1 m)=554 nm

The wavelength of radiation from excited state 1 to ground state can be calculated as,

λ1−0=h×cE1−E0=6.626× 10 −34 J⋅s×3× 108 m/s( 2.11 eV×( 1.6× 10 −19 J 1 eV )−0)=558×10−9 m( 10 9 nm 1 m)=558 nm

Conclusion:

Therefore, the wavelength at excitation state of 4.35 eV is 285 nm , the wavelengths of resonance fluorescence at excited state are 554 nm and 558 nm

Answer 14

(b)

To determine

The wavelength at excitation state of 4.35 eV and the wavelength of radiation at excited states.

Answer 15

(b)

Expert Solution

Answer to Problem 67P

The wavelength at excitation state of 4.35 eV is 285 nm , the wavelengths of resonance fluorescence at excited state are 554 nm and 558 nm

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation:

The wavelength of radiation from excited state 3 to state 1 can be calculated as,

λ3−1=h×cE3−E1=6.626× 10 −34 J⋅s×3× 108 m/s( 4.35 eV×( 1.6× 10 −19 J 1 eV )−2.11 eV×( 1.6× 10 −19 J 1 eV ))=554×10−9 m( 10 9 nm 1 m)=554 nm

The wavelength of radiation from excited state 1 to ground state can be calculated as,

λ1−0=h×cE1−E0=6.626× 10 −34 J⋅s×3× 108 m/s( 2.11 eV×( 1.6× 10 −19 J 1 eV )−0)=558×10−9 m( 10 9 nm 1 m)=558 nm

Conclusion:

Therefore, the wavelength at excitation state of 4.35 eV is 285 nm , the wavelengths of resonance fluorescence at excited state are 554 nm and 558 nm

Answer 20

Ch. 31 - Prob. 1P Ch. 31 - Prob. 2P Ch. 31 - Prob. 3P Ch. 31 - Prob. 4P Ch. 31 - Prob. 5P Ch. 31 - Prob. 6P Ch. 31 - Prob. 7P Ch. 31 - Prob. 8P Ch. 31 - Prob. 9P Ch. 31 - Prob. 10P

The maximum wavelength of radiation .

Concept explainers

The maximum wavelength of radiation.

Answer to Problem 67P

Explanation of Solution

The wavelength at excitation state of 4.35 eV and the wavelength of radiation at excited states.

Answer to Problem 67P

Explanation of Solution

Want to see more full solutions like this?

Chapter 31 Solutions