Frequency of the first five lines in Lyman series has to be calculated.

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Answer 1

Question

Chapter 4, Problem 4.41P

Interpretation Introduction

Interpretation:

Frequency of the first five lines in Lyman series has to be calculated.

Expert Solution & Answer

Explanation of Solution

Frequency of First line:

Energy of photon emitted on relaxation from n=2 state to n=1 state in first line of Lyman series and is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(112−122)=(2.1799×10−18 J)(11−14)=1.6349×10−18 J

Frequency of photon is calculated as shown below.

E=hvv=Eh=1.6349×10−18 J6.626×10−34 J⋅s=0.246740114699668×1016 s-1=2.467×1015 s−1

Therefore, the frequency of photon is calculated as 2.467×1015 s−1.

Frequency of Second line:

Energy of photon emitted on relaxation from n=3 state to n=1 state in second line of Lyman series and 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

Frequency of photon is calculated as shown below.

E=hvv=Eh=1.9377×10−18 J6.626×10−34 J⋅s=0.2924388771506188×1016 s-1=2.924×1015 s−1

Therefore, the frequency of photon is calculated as 2.924×1015 s−1.

Frequency of Third line:

Energy of photon emitted on relaxation from n=4 state to n=1 state in third line of Lyman series and is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(112−142)=(2.1799×10−18 J)(11−116)=2.0436×10−18 J

Frequency of photon is calculated as shown below.

E=hvv=Eh=2.0436×10−18 J6.626×10−34 J⋅s=0.3084213703591911×1016 s-1=3.08×1015 s−1

Therefore, the frequency of photon is calculated as 3.08×1015 s−1.

Frequency of Fourth line:

Energy of photon emitted on relaxation from n=5 state to n=1 state in fourth line of Lyman series and is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(112−152)=(2.1799×10−18 J)(11−125)=2.0927×10−18 J

Frequency of photon is calculated as shown below.

E=hvv=Eh=2.0927×10−18 J6.626×10−34 J⋅s=0.3158315725928162×1016 s-1=3.16×1015 s−1

Therefore, the frequency of photon is calculated as 3.16×1015 s−1.

Frequency of Fifth line:

Energy of photon emitted on relaxation from n=6 state to n=1 state in fifth line of Lyman series and is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(112−162)=(2.1799×10−18 J)(11−136)=2.1193×10−18 J

Frequency of photon is calculated as shown below.

E=hvv=Eh=2.1193×10−18 J6.626×10−34 J⋅s=0.3198460609719288×1016 s-1=3.19×1015 s−1

Therefore, the frequency of photon is calculated as 3.19×1015 s−1.

Summary of the frequency of first five lines in Lyman series is given below.

ni	2	3	4	5	6
Frequency/×1015 s−1	2.467	2.924	3.08	3.16	3.19

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Answer 2

Question

Chapter 4, Problem 4.41P

Interpretation Introduction

Interpretation:

Frequency of the first five lines in Lyman series has to be calculated.

Expert Solution & Answer

Explanation of Solution

Frequency of First line:

Energy of photon emitted on relaxation from n=2 state to n=1 state in first line of Lyman series and is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(112−122)=(2.1799×10−18 J)(11−14)=1.6349×10−18 J

Frequency of photon is calculated as shown below.

E=hvv=Eh=1.6349×10−18 J6.626×10−34 J⋅s=0.246740114699668×1016 s-1=2.467×1015 s−1

Therefore, the frequency of photon is calculated as 2.467×1015 s−1.

Frequency of Second line:

Energy of photon emitted on relaxation from n=3 state to n=1 state in second line of Lyman series and 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

Frequency of photon is calculated as shown below.

E=hvv=Eh=1.9377×10−18 J6.626×10−34 J⋅s=0.2924388771506188×1016 s-1=2.924×1015 s−1

Therefore, the frequency of photon is calculated as 2.924×1015 s−1.

Frequency of Third line:

Energy of photon emitted on relaxation from n=4 state to n=1 state in third line of Lyman series and is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(112−142)=(2.1799×10−18 J)(11−116)=2.0436×10−18 J

Frequency of photon is calculated as shown below.

E=hvv=Eh=2.0436×10−18 J6.626×10−34 J⋅s=0.3084213703591911×1016 s-1=3.08×1015 s−1

Therefore, the frequency of photon is calculated as 3.08×1015 s−1.

Frequency of Fourth line:

Energy of photon emitted on relaxation from n=5 state to n=1 state in fourth line of Lyman series and is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(112−152)=(2.1799×10−18 J)(11−125)=2.0927×10−18 J

Frequency of photon is calculated as shown below.

E=hvv=Eh=2.0927×10−18 J6.626×10−34 J⋅s=0.3158315725928162×1016 s-1=3.16×1015 s−1

Therefore, the frequency of photon is calculated as 3.16×1015 s−1.

Frequency of Fifth line:

Energy of photon emitted on relaxation from n=6 state to n=1 state in fifth line of Lyman series and is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(112−162)=(2.1799×10−18 J)(11−136)=2.1193×10−18 J

Frequency of photon is calculated as shown below.

E=hvv=Eh=2.1193×10−18 J6.626×10−34 J⋅s=0.3198460609719288×1016 s-1=3.19×1015 s−1

Therefore, the frequency of photon is calculated as 3.19×1015 s−1.

Summary of the frequency of first five lines in Lyman series is given below.

ni	2	3	4	5	6
Frequency/×1015 s−1	2.467	2.924	3.08	3.16	3.19

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Answer 3

Question

Chapter 4, Problem 4.41P

Interpretation Introduction

Interpretation:

Frequency of the first five lines in Lyman series has to be calculated.

Expert Solution & Answer

Explanation of Solution

Frequency of First line:

Energy of photon emitted on relaxation from n=2 state to n=1 state in first line of Lyman series and is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(112−122)=(2.1799×10−18 J)(11−14)=1.6349×10−18 J

Frequency of photon is calculated as shown below.

E=hvv=Eh=1.6349×10−18 J6.626×10−34 J⋅s=0.246740114699668×1016 s-1=2.467×1015 s−1

Therefore, the frequency of photon is calculated as 2.467×1015 s−1.

Frequency of Second line:

Energy of photon emitted on relaxation from n=3 state to n=1 state in second line of Lyman series and 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

Frequency of photon is calculated as shown below.

E=hvv=Eh=1.9377×10−18 J6.626×10−34 J⋅s=0.2924388771506188×1016 s-1=2.924×1015 s−1

Therefore, the frequency of photon is calculated as 2.924×1015 s−1.

Frequency of Third line:

Energy of photon emitted on relaxation from n=4 state to n=1 state in third line of Lyman series and is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(112−142)=(2.1799×10−18 J)(11−116)=2.0436×10−18 J

Frequency of photon is calculated as shown below.

E=hvv=Eh=2.0436×10−18 J6.626×10−34 J⋅s=0.3084213703591911×1016 s-1=3.08×1015 s−1

Therefore, the frequency of photon is calculated as 3.08×1015 s−1.

Frequency of Fourth line:

Energy of photon emitted on relaxation from n=5 state to n=1 state in fourth line of Lyman series and is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(112−152)=(2.1799×10−18 J)(11−125)=2.0927×10−18 J

Frequency of photon is calculated as shown below.

E=hvv=Eh=2.0927×10−18 J6.626×10−34 J⋅s=0.3158315725928162×1016 s-1=3.16×1015 s−1

Therefore, the frequency of photon is calculated as 3.16×1015 s−1.

Frequency of Fifth line:

Energy of photon emitted on relaxation from n=6 state to n=1 state in fifth line of Lyman series and is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(112−162)=(2.1799×10−18 J)(11−136)=2.1193×10−18 J

Frequency of photon is calculated as shown below.

E=hvv=Eh=2.1193×10−18 J6.626×10−34 J⋅s=0.3198460609719288×1016 s-1=3.19×1015 s−1

Therefore, the frequency of photon is calculated as 3.19×1015 s−1.

Summary of the frequency of first five lines in Lyman series is given below.

ni	2	3	4	5	6
Frequency/×1015 s−1	2.467	2.924	3.08	3.16	3.19

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Answer 4

Question

Chapter 4, Problem 4.41P

Interpretation Introduction

Interpretation:

Frequency of the first five lines in Lyman series has to be calculated.

Expert Solution & Answer

Explanation of Solution

Frequency of First line:

Energy of photon emitted on relaxation from n=2 state to n=1 state in first line of Lyman series and is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(112−122)=(2.1799×10−18 J)(11−14)=1.6349×10−18 J

Frequency of photon is calculated as shown below.

E=hvv=Eh=1.6349×10−18 J6.626×10−34 J⋅s=0.246740114699668×1016 s-1=2.467×1015 s−1

Therefore, the frequency of photon is calculated as 2.467×1015 s−1.

Frequency of Second line:

Energy of photon emitted on relaxation from n=3 state to n=1 state in second line of Lyman series and 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

Frequency of photon is calculated as shown below.

E=hvv=Eh=1.9377×10−18 J6.626×10−34 J⋅s=0.2924388771506188×1016 s-1=2.924×1015 s−1

Therefore, the frequency of photon is calculated as 2.924×1015 s−1.

Frequency of Third line:

Energy of photon emitted on relaxation from n=4 state to n=1 state in third line of Lyman series and is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(112−142)=(2.1799×10−18 J)(11−116)=2.0436×10−18 J

Frequency of photon is calculated as shown below.

E=hvv=Eh=2.0436×10−18 J6.626×10−34 J⋅s=0.3084213703591911×1016 s-1=3.08×1015 s−1

Therefore, the frequency of photon is calculated as 3.08×1015 s−1.

Frequency of Fourth line:

Energy of photon emitted on relaxation from n=5 state to n=1 state in fourth line of Lyman series and is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(112−152)=(2.1799×10−18 J)(11−125)=2.0927×10−18 J

Frequency of photon is calculated as shown below.

E=hvv=Eh=2.0927×10−18 J6.626×10−34 J⋅s=0.3158315725928162×1016 s-1=3.16×1015 s−1

Therefore, the frequency of photon is calculated as 3.16×1015 s−1.

Frequency of Fifth line:

Energy of photon emitted on relaxation from n=6 state to n=1 state in fifth line of Lyman series and is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(112−162)=(2.1799×10−18 J)(11−136)=2.1193×10−18 J

Frequency of photon is calculated as shown below.

E=hvv=Eh=2.1193×10−18 J6.626×10−34 J⋅s=0.3198460609719288×1016 s-1=3.19×1015 s−1

Therefore, the frequency of photon is calculated as 3.19×1015 s−1.

Summary of the frequency of first five lines in Lyman series is given below.

ni	2	3	4	5	6
Frequency/×1015 s−1	2.467	2.924	3.08	3.16	3.19

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Answer 5

Chapter 4, Problem 4.41P

Interpretation Introduction

Interpretation:

Frequency of the first five lines in Lyman series has to be calculated.

Expert Solution & Answer

Explanation of Solution

Frequency of First line:

Energy of photon emitted on relaxation from n=2 state to n=1 state in first line of Lyman series and is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(112−122)=(2.1799×10−18 J)(11−14)=1.6349×10−18 J

Frequency of photon is calculated as shown below.

E=hvv=Eh=1.6349×10−18 J6.626×10−34 J⋅s=0.246740114699668×1016 s-1=2.467×1015 s−1

Therefore, the frequency of photon is calculated as 2.467×1015 s−1.

Frequency of Second line:

Energy of photon emitted on relaxation from n=3 state to n=1 state in second line of Lyman series and 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

Frequency of photon is calculated as shown below.

E=hvv=Eh=1.9377×10−18 J6.626×10−34 J⋅s=0.2924388771506188×1016 s-1=2.924×1015 s−1

Therefore, the frequency of photon is calculated as 2.924×1015 s−1.

Frequency of Third line:

Energy of photon emitted on relaxation from n=4 state to n=1 state in third line of Lyman series and is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(112−142)=(2.1799×10−18 J)(11−116)=2.0436×10−18 J

Frequency of photon is calculated as shown below.

E=hvv=Eh=2.0436×10−18 J6.626×10−34 J⋅s=0.3084213703591911×1016 s-1=3.08×1015 s−1

Therefore, the frequency of photon is calculated as 3.08×1015 s−1.

Frequency of Fourth line:

Energy of photon emitted on relaxation from n=5 state to n=1 state in fourth line of Lyman series and is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(112−152)=(2.1799×10−18 J)(11−125)=2.0927×10−18 J

Frequency of photon is calculated as shown below.

E=hvv=Eh=2.0927×10−18 J6.626×10−34 J⋅s=0.3158315725928162×1016 s-1=3.16×1015 s−1

Therefore, the frequency of photon is calculated as 3.16×1015 s−1.

Frequency of Fifth line:

Energy of photon emitted on relaxation from n=6 state to n=1 state in fifth line of Lyman series and is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(112−162)=(2.1799×10−18 J)(11−136)=2.1193×10−18 J

Frequency of photon is calculated as shown below.

E=hvv=Eh=2.1193×10−18 J6.626×10−34 J⋅s=0.3198460609719288×1016 s-1=3.19×1015 s−1

Therefore, the frequency of photon is calculated as 3.19×1015 s−1.

Summary of the frequency of first five lines in Lyman series is given below.

ni	2	3	4	5	6
Frequency/×1015 s−1	2.467	2.924	3.08	3.16	3.19

Answer 6

Chapter 4, Problem 4.41P

Interpretation Introduction

Interpretation:

Frequency of the first five lines in Lyman series has to be calculated.

Expert Solution & Answer

Explanation of Solution

Frequency of First line:

Energy of photon emitted on relaxation from n=2 state to n=1 state in first line of Lyman series and is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(112−122)=(2.1799×10−18 J)(11−14)=1.6349×10−18 J

Frequency of photon is calculated as shown below.

E=hvv=Eh=1.6349×10−18 J6.626×10−34 J⋅s=0.246740114699668×1016 s-1=2.467×1015 s−1

Therefore, the frequency of photon is calculated as 2.467×1015 s−1.

Frequency of Second line:

Energy of photon emitted on relaxation from n=3 state to n=1 state in second line of Lyman series and 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

Frequency of photon is calculated as shown below.

E=hvv=Eh=1.9377×10−18 J6.626×10−34 J⋅s=0.2924388771506188×1016 s-1=2.924×1015 s−1

Therefore, the frequency of photon is calculated as 2.924×1015 s−1.

Frequency of Third line:

Energy of photon emitted on relaxation from n=4 state to n=1 state in third line of Lyman series and is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(112−142)=(2.1799×10−18 J)(11−116)=2.0436×10−18 J

Frequency of photon is calculated as shown below.

E=hvv=Eh=2.0436×10−18 J6.626×10−34 J⋅s=0.3084213703591911×1016 s-1=3.08×1015 s−1

Therefore, the frequency of photon is calculated as 3.08×1015 s−1.

Frequency of Fourth line:

Energy of photon emitted on relaxation from n=5 state to n=1 state in fourth line of Lyman series and is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(112−152)=(2.1799×10−18 J)(11−125)=2.0927×10−18 J

Frequency of photon is calculated as shown below.

E=hvv=Eh=2.0927×10−18 J6.626×10−34 J⋅s=0.3158315725928162×1016 s-1=3.16×1015 s−1

Therefore, the frequency of photon is calculated as 3.16×1015 s−1.

Frequency of Fifth line:

Energy of photon emitted on relaxation from n=6 state to n=1 state in fifth line of Lyman series and is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(112−162)=(2.1799×10−18 J)(11−136)=2.1193×10−18 J

Frequency of photon is calculated as shown below.

E=hvv=Eh=2.1193×10−18 J6.626×10−34 J⋅s=0.3198460609719288×1016 s-1=3.19×1015 s−1

Therefore, the frequency of photon is calculated as 3.19×1015 s−1.

Summary of the frequency of first five lines in Lyman series is given below.

ni	2	3	4	5	6
Frequency/×1015 s−1	2.467	2.924	3.08	3.16	3.19

Answer 7

Chapter 4, Problem 4.41P

Interpretation Introduction

Interpretation:

Frequency of the first five lines in Lyman series has to be calculated.

Answer 8

Expert Solution & Answer

Explanation of Solution

Frequency of First line:

Energy of photon emitted on relaxation from n=2 state to n=1 state in first line of Lyman series and is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(112−122)=(2.1799×10−18 J)(11−14)=1.6349×10−18 J

Frequency of photon is calculated as shown below.

E=hvv=Eh=1.6349×10−18 J6.626×10−34 J⋅s=0.246740114699668×1016 s-1=2.467×1015 s−1

Therefore, the frequency of photon is calculated as 2.467×1015 s−1.

Frequency of Second line:

Energy of photon emitted on relaxation from n=3 state to n=1 state in second line of Lyman series and 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

Frequency of photon is calculated as shown below.

E=hvv=Eh=1.9377×10−18 J6.626×10−34 J⋅s=0.2924388771506188×1016 s-1=2.924×1015 s−1

Therefore, the frequency of photon is calculated as 2.924×1015 s−1.

Frequency of Third line:

Energy of photon emitted on relaxation from n=4 state to n=1 state in third line of Lyman series and is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(112−142)=(2.1799×10−18 J)(11−116)=2.0436×10−18 J

Frequency of photon is calculated as shown below.

E=hvv=Eh=2.0436×10−18 J6.626×10−34 J⋅s=0.3084213703591911×1016 s-1=3.08×1015 s−1

Therefore, the frequency of photon is calculated as 3.08×1015 s−1.

Frequency of Fourth line:

Energy of photon emitted on relaxation from n=5 state to n=1 state in fourth line of Lyman series and is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(112−152)=(2.1799×10−18 J)(11−125)=2.0927×10−18 J

Frequency of photon is calculated as shown below.

E=hvv=Eh=2.0927×10−18 J6.626×10−34 J⋅s=0.3158315725928162×1016 s-1=3.16×1015 s−1

Therefore, the frequency of photon is calculated as 3.16×1015 s−1.

Frequency of Fifth line:

Energy of photon emitted on relaxation from n=6 state to n=1 state in fifth line of Lyman series and is calculated as shown below.

Ephoton=(2.1799×10−18 J)(1nf2−1ni2)=(2.1799×10−18 J)(112−162)=(2.1799×10−18 J)(11−136)=2.1193×10−18 J

Frequency of photon is calculated as shown below.

E=hvv=Eh=2.1193×10−18 J6.626×10−34 J⋅s=0.3198460609719288×1016 s-1=3.19×1015 s−1

Therefore, the frequency of photon is calculated as 3.19×1015 s−1.

Summary of the frequency of first five lines in Lyman series is given below.

ni	2	3	4	5	6
Frequency/×1015 s−1	2.467	2.924	3.08	3.16	3.19

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Ch. 4 - Prob. 4.1P Ch. 4 - Prob. 4.2P Ch. 4 - Prob. 4.3P Ch. 4 - Prob. 4.4P Ch. 4 - Prob. 4.5P Ch. 4 - Prob. 4.6P Ch. 4 - Prob. 4.7P Ch. 4 - Prob. 4.8P Ch. 4 - Prob. 4.9P Ch. 4 - Prob. 4.10P

Frequency of the first five lines in Lyman series has to be calculated.

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Interpretation:

Frequency of the first five lines in Lyman series has to be calculated.

Explanation of Solution

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Chapter 4 Solutions

Frequency of the first five lines in Lyman series has to be calculated.

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Interpretation: Frequency of the first five lines in Lyman series has to be calculated.

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Chapter 4 Solutions

Interpretation:

Frequency of the first five lines in Lyman series has to be calculated.