Wavelength of the light that is emitted will be greater or not if the electron relax directly back to n = 1 state has to be given. Concept Introduction: By absorbing or emitting electromagnetic radiations, the atom undergoes transition from one stationary state to another. If transition occurs from a high energy state to a lower energy state, then it is accompanied by emission of a photon for hydrogen atom. Energy of photon that is emitted can be calculated using the equation given below. E p h o t o n = ( 2.1799 × 10 − 18 J ) ( 1 n f 2 − 1 n i 2 ) Where, n f is the final energy state. n i is the initial energy state.

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Answer 1

Question

Chapter 4, Problem 4.37P

Interpretation Introduction

Interpretation:

Wavelength of the light that is emitted will be greater or not if the electron relax directly back to n=1 state has to be given.

Concept Introduction:

By absorbing or emitting electromagnetic radiations, the atom undergoes transition from one stationary state to another. If transition occurs from a high energy state to a lower energy state, then it is accompanied by emission of a photon for hydrogen atom. Energy of photon that is emitted can be calculated using the equation given below.

    Ephoton=(2.1799×10−18 J)(1nf2−1ni2)

Where,

    nf is the final energy state.

    ni is the initial energy state.

Expert Solution & Answer

Explanation of Solution

Hydrogen atom absorbs photon and is excited to n=3 state. This state is known as second excited state because n=3.

Energy of photon emitted on relaxation from n=3 state to n=2 state is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(122−132)=(2.1799×10−18 J)(14−19)=0.3027×10−18 J

Wavelength of the light is calculated as shown below.

λ1=hcE=6.626×10−34 J⋅s×2.9979×108 m⋅s−10.3027×10−18 J=65.62×10−8 m=656 nm

Wavelength of light when photon is emitted on relaxation from n=3 to n=2 state is 656 nm.

Energy of photon emitted on relaxation from n=3 state to n=1 state is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(112−132)=(2.1799×10−18 J)(11−19)=1.9377×10−18 J

Wavelength of the light is calculated as shown below.

λ2=hcE=6.626×10−34 J⋅s×2.9979×108 m⋅s−11.9377×10−18 J=10.25×10−8 m=103 nm

Wavelength of light when photon is emitted on relaxation from n=3 to n=1 state is 103 nm.

Therefore, comparing both wavelengths of light, it is found that the wavelength of light emitted from n=3 to n=2 is greater than for the wavelength of light emitted from n=3 to n=1.

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Answer 2

Question

Chapter 4, Problem 4.37P

Interpretation Introduction

Interpretation:

Wavelength of the light that is emitted will be greater or not if the electron relax directly back to n=1 state has to be given.

Concept Introduction:

By absorbing or emitting electromagnetic radiations, the atom undergoes transition from one stationary state to another. If transition occurs from a high energy state to a lower energy state, then it is accompanied by emission of a photon for hydrogen atom. Energy of photon that is emitted can be calculated using the equation given below.

    Ephoton=(2.1799×10−18 J)(1nf2−1ni2)

Where,

    nf is the final energy state.

    ni is the initial energy state.

Expert Solution & Answer

Explanation of Solution

Hydrogen atom absorbs photon and is excited to n=3 state. This state is known as second excited state because n=3.

Energy of photon emitted on relaxation from n=3 state to n=2 state is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(122−132)=(2.1799×10−18 J)(14−19)=0.3027×10−18 J

Wavelength of the light is calculated as shown below.

λ1=hcE=6.626×10−34 J⋅s×2.9979×108 m⋅s−10.3027×10−18 J=65.62×10−8 m=656 nm

Wavelength of light when photon is emitted on relaxation from n=3 to n=2 state is 656 nm.

Energy of photon emitted on relaxation from n=3 state to n=1 state is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(112−132)=(2.1799×10−18 J)(11−19)=1.9377×10−18 J

Wavelength of the light is calculated as shown below.

λ2=hcE=6.626×10−34 J⋅s×2.9979×108 m⋅s−11.9377×10−18 J=10.25×10−8 m=103 nm

Wavelength of light when photon is emitted on relaxation from n=3 to n=1 state is 103 nm.

Therefore, comparing both wavelengths of light, it is found that the wavelength of light emitted from n=3 to n=2 is greater than for the wavelength of light emitted from n=3 to n=1.

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Answer 3

Question

Chapter 4, Problem 4.37P

Interpretation Introduction

Interpretation:

Wavelength of the light that is emitted will be greater or not if the electron relax directly back to n=1 state has to be given.

Concept Introduction:

By absorbing or emitting electromagnetic radiations, the atom undergoes transition from one stationary state to another. If transition occurs from a high energy state to a lower energy state, then it is accompanied by emission of a photon for hydrogen atom. Energy of photon that is emitted can be calculated using the equation given below.

    Ephoton=(2.1799×10−18 J)(1nf2−1ni2)

Where,

    nf is the final energy state.

    ni is the initial energy state.

Expert Solution & Answer

Explanation of Solution

Hydrogen atom absorbs photon and is excited to n=3 state. This state is known as second excited state because n=3.

Energy of photon emitted on relaxation from n=3 state to n=2 state is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(122−132)=(2.1799×10−18 J)(14−19)=0.3027×10−18 J

Wavelength of the light is calculated as shown below.

λ1=hcE=6.626×10−34 J⋅s×2.9979×108 m⋅s−10.3027×10−18 J=65.62×10−8 m=656 nm

Wavelength of light when photon is emitted on relaxation from n=3 to n=2 state is 656 nm.

Energy of photon emitted on relaxation from n=3 state to n=1 state is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(112−132)=(2.1799×10−18 J)(11−19)=1.9377×10−18 J

Wavelength of the light is calculated as shown below.

λ2=hcE=6.626×10−34 J⋅s×2.9979×108 m⋅s−11.9377×10−18 J=10.25×10−8 m=103 nm

Wavelength of light when photon is emitted on relaxation from n=3 to n=1 state is 103 nm.

Therefore, comparing both wavelengths of light, it is found that the wavelength of light emitted from n=3 to n=2 is greater than for the wavelength of light emitted from n=3 to n=1.

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Answer 4

Question

Chapter 4, Problem 4.37P

Interpretation Introduction

Interpretation:

Wavelength of the light that is emitted will be greater or not if the electron relax directly back to n=1 state has to be given.

Concept Introduction:

By absorbing or emitting electromagnetic radiations, the atom undergoes transition from one stationary state to another. If transition occurs from a high energy state to a lower energy state, then it is accompanied by emission of a photon for hydrogen atom. Energy of photon that is emitted can be calculated using the equation given below.

    Ephoton=(2.1799×10−18 J)(1nf2−1ni2)

Where,

    nf is the final energy state.

    ni is the initial energy state.

Expert Solution & Answer

Explanation of Solution

Hydrogen atom absorbs photon and is excited to n=3 state. This state is known as second excited state because n=3.

Energy of photon emitted on relaxation from n=3 state to n=2 state is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(122−132)=(2.1799×10−18 J)(14−19)=0.3027×10−18 J

Wavelength of the light is calculated as shown below.

λ1=hcE=6.626×10−34 J⋅s×2.9979×108 m⋅s−10.3027×10−18 J=65.62×10−8 m=656 nm

Wavelength of light when photon is emitted on relaxation from n=3 to n=2 state is 656 nm.

Energy of photon emitted on relaxation from n=3 state to n=1 state is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(112−132)=(2.1799×10−18 J)(11−19)=1.9377×10−18 J

Wavelength of the light is calculated as shown below.

λ2=hcE=6.626×10−34 J⋅s×2.9979×108 m⋅s−11.9377×10−18 J=10.25×10−8 m=103 nm

Wavelength of light when photon is emitted on relaxation from n=3 to n=1 state is 103 nm.

Therefore, comparing both wavelengths of light, it is found that the wavelength of light emitted from n=3 to n=2 is greater than for the wavelength of light emitted from n=3 to n=1.

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Answer 5

Chapter 4, Problem 4.37P

Interpretation Introduction

Interpretation:

Wavelength of the light that is emitted will be greater or not if the electron relax directly back to n=1 state has to be given.

Concept Introduction:

By absorbing or emitting electromagnetic radiations, the atom undergoes transition from one stationary state to another. If transition occurs from a high energy state to a lower energy state, then it is accompanied by emission of a photon for hydrogen atom. Energy of photon that is emitted can be calculated using the equation given below.

    Ephoton=(2.1799×10−18 J)(1nf2−1ni2)

Where,

    nf is the final energy state.

    ni is the initial energy state.

Expert Solution & Answer

Explanation of Solution

Hydrogen atom absorbs photon and is excited to n=3 state. This state is known as second excited state because n=3.

Energy of photon emitted on relaxation from n=3 state to n=2 state is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(122−132)=(2.1799×10−18 J)(14−19)=0.3027×10−18 J

Wavelength of the light is calculated as shown below.

λ1=hcE=6.626×10−34 J⋅s×2.9979×108 m⋅s−10.3027×10−18 J=65.62×10−8 m=656 nm

Wavelength of light when photon is emitted on relaxation from n=3 to n=2 state is 656 nm.

Energy of photon emitted on relaxation from n=3 state to n=1 state is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(112−132)=(2.1799×10−18 J)(11−19)=1.9377×10−18 J

Wavelength of the light is calculated as shown below.

λ2=hcE=6.626×10−34 J⋅s×2.9979×108 m⋅s−11.9377×10−18 J=10.25×10−8 m=103 nm

Wavelength of light when photon is emitted on relaxation from n=3 to n=1 state is 103 nm.

Therefore, comparing both wavelengths of light, it is found that the wavelength of light emitted from n=3 to n=2 is greater than for the wavelength of light emitted from n=3 to n=1.

Answer 6

Chapter 4, Problem 4.37P

Interpretation Introduction

Interpretation:

Wavelength of the light that is emitted will be greater or not if the electron relax directly back to n=1 state has to be given.

Concept Introduction:

By absorbing or emitting electromagnetic radiations, the atom undergoes transition from one stationary state to another. If transition occurs from a high energy state to a lower energy state, then it is accompanied by emission of a photon for hydrogen atom. Energy of photon that is emitted can be calculated using the equation given below.

    Ephoton=(2.1799×10−18 J)(1nf2−1ni2)

Where,

    nf is the final energy state.

    ni is the initial energy state.

Expert Solution & Answer

Explanation of Solution

Hydrogen atom absorbs photon and is excited to n=3 state. This state is known as second excited state because n=3.

Energy of photon emitted on relaxation from n=3 state to n=2 state is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(122−132)=(2.1799×10−18 J)(14−19)=0.3027×10−18 J

Wavelength of the light is calculated as shown below.

λ1=hcE=6.626×10−34 J⋅s×2.9979×108 m⋅s−10.3027×10−18 J=65.62×10−8 m=656 nm

Wavelength of light when photon is emitted on relaxation from n=3 to n=2 state is 656 nm.

Energy of photon emitted on relaxation from n=3 state to n=1 state is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(112−132)=(2.1799×10−18 J)(11−19)=1.9377×10−18 J

Wavelength of the light is calculated as shown below.

λ2=hcE=6.626×10−34 J⋅s×2.9979×108 m⋅s−11.9377×10−18 J=10.25×10−8 m=103 nm

Wavelength of light when photon is emitted on relaxation from n=3 to n=1 state is 103 nm.

Therefore, comparing both wavelengths of light, it is found that the wavelength of light emitted from n=3 to n=2 is greater than for the wavelength of light emitted from n=3 to n=1.

Answer 7

Chapter 4, Problem 4.37P

Interpretation Introduction

Interpretation:

Wavelength of the light that is emitted will be greater or not if the electron relax directly back to n=1 state has to be given.

Concept Introduction:

By absorbing or emitting electromagnetic radiations, the atom undergoes transition from one stationary state to another. If transition occurs from a high energy state to a lower energy state, then it is accompanied by emission of a photon for hydrogen atom. Energy of photon that is emitted can be calculated using the equation given below.

    Ephoton=(2.1799×10−18 J)(1nf2−1ni2)

Where,

    nf is the final energy state.

    ni is the initial energy state.

Answer 8

Expert Solution & Answer

Explanation of Solution

Hydrogen atom absorbs photon and is excited to n=3 state. This state is known as second excited state because n=3.

Energy of photon emitted on relaxation from n=3 state to n=2 state is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(122−132)=(2.1799×10−18 J)(14−19)=0.3027×10−18 J

Wavelength of the light is calculated as shown below.

λ1=hcE=6.626×10−34 J⋅s×2.9979×108 m⋅s−10.3027×10−18 J=65.62×10−8 m=656 nm

Wavelength of light when photon is emitted on relaxation from n=3 to n=2 state is 656 nm.

Energy of photon emitted on relaxation from n=3 state to n=1 state is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(112−132)=(2.1799×10−18 J)(11−19)=1.9377×10−18 J

Wavelength of the light is calculated as shown below.

λ2=hcE=6.626×10−34 J⋅s×2.9979×108 m⋅s−11.9377×10−18 J=10.25×10−8 m=103 nm

Wavelength of light when photon is emitted on relaxation from n=3 to n=1 state is 103 nm.

Therefore, comparing both wavelengths of light, it is found that the wavelength of light emitted from n=3 to n=2 is greater than for the wavelength of light emitted from n=3 to n=1.