The wavelength of the next line in the given series of lines of spectrum of atomic hydrogen has to be identified.

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Answer 1

Question

Chapter 8, Problem 8.4P

(a)

Interpretation Introduction

Interpretation:

The wavelength of the next line in the given series of lines of spectrum of atomic hydrogen has to be identified.

(a)

Expert Solution

Explanation of Solution

All the lines in the hydrogen spectrum suits the Rydberg formula.

$1λ = RH(1n12 - 1n22)Where λ is the wavelength of the emitted photonR is Rydberg’s constantZ is the atomic number (1 for H, 2 for He)n1 and n2 are the principal quantum numbers for the initial and final states$

Find $n1$ from the value of $λmax$ , which arises from the transition of $n1+ 1 → n1$ .

$1λmaxRH = 1n12 - (1n1 + 1)2 = 2n1 + 1n12 (n1 + 1)2λmaxRH = n12 (n1 + 1)22n1 + 1 = (656.46 × 10-9m) × (109677× 102m-1 ) = 7.20hence n1 = 2, as determined by trial and error substitution. Therefore,the transitions are given as,υ = 1λ = (109677 cm-1) × (14 - 1n22) n2 = 3,4,5,6Next line has n2 = 7υ = 1λ = (109677 cm-1) × (14 - 149) = 397.13nm$

The energy needed to ionize the atom is obtained by allowing $n2→ ∞$

$υ = 1λ∞ = (109677 cm-1) × (14 - 0) = 27419cm-1υ = 3.40ev$

(b)

Interpretation Introduction

Interpretation:

The ionization energy of hydrogen atom when it is in lower state of transition has to be identified.

(b)

Expert Solution

Explanation of Solution

The energy needed to ionize the atom is obtained by allowing $n2→ ∞$

$υ = 1λ∞ = (109677 cm-1) × (14 - 0) = 27419cm-11cm-1 = 1.2398 ×10-4evυ = 3.40ev$

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Answer 2

Question

Chapter 8, Problem 8.4P

(a)

Interpretation Introduction

Interpretation:

The wavelength of the next line in the given series of lines of spectrum of atomic hydrogen has to be identified.

(a)

Expert Solution

Explanation of Solution

All the lines in the hydrogen spectrum suits the Rydberg formula.

$1λ = RH(1n12 - 1n22)Where λ is the wavelength of the emitted photonR is Rydberg’s constantZ is the atomic number (1 for H, 2 for He)n1 and n2 are the principal quantum numbers for the initial and final states$

Find $n1$ from the value of $λmax$ , which arises from the transition of $n1+ 1 → n1$ .

$1λmaxRH = 1n12 - (1n1 + 1)2 = 2n1 + 1n12 (n1 + 1)2λmaxRH = n12 (n1 + 1)22n1 + 1 = (656.46 × 10-9m) × (109677× 102m-1 ) = 7.20hence n1 = 2, as determined by trial and error substitution. Therefore,the transitions are given as,υ = 1λ = (109677 cm-1) × (14 - 1n22) n2 = 3,4,5,6Next line has n2 = 7υ = 1λ = (109677 cm-1) × (14 - 149) = 397.13nm$

The energy needed to ionize the atom is obtained by allowing $n2→ ∞$

$υ = 1λ∞ = (109677 cm-1) × (14 - 0) = 27419cm-1υ = 3.40ev$

(b)

Interpretation Introduction

Interpretation:

The ionization energy of hydrogen atom when it is in lower state of transition has to be identified.

(b)

Expert Solution

Explanation of Solution

The energy needed to ionize the atom is obtained by allowing $n2→ ∞$

$υ = 1λ∞ = (109677 cm-1) × (14 - 0) = 27419cm-11cm-1 = 1.2398 ×10-4evυ = 3.40ev$

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Answer 3

Question

Chapter 8, Problem 8.4P

(a)

Interpretation Introduction

Interpretation:

The wavelength of the next line in the given series of lines of spectrum of atomic hydrogen has to be identified.

(a)

Expert Solution

Explanation of Solution

All the lines in the hydrogen spectrum suits the Rydberg formula.

$1λ = RH(1n12 - 1n22)Where λ is the wavelength of the emitted photonR is Rydberg’s constantZ is the atomic number (1 for H, 2 for He)n1 and n2 are the principal quantum numbers for the initial and final states$

Find $n1$ from the value of $λmax$ , which arises from the transition of $n1+ 1 → n1$ .

$1λmaxRH = 1n12 - (1n1 + 1)2 = 2n1 + 1n12 (n1 + 1)2λmaxRH = n12 (n1 + 1)22n1 + 1 = (656.46 × 10-9m) × (109677× 102m-1 ) = 7.20hence n1 = 2, as determined by trial and error substitution. Therefore,the transitions are given as,υ = 1λ = (109677 cm-1) × (14 - 1n22) n2 = 3,4,5,6Next line has n2 = 7υ = 1λ = (109677 cm-1) × (14 - 149) = 397.13nm$

The energy needed to ionize the atom is obtained by allowing $n2→ ∞$

$υ = 1λ∞ = (109677 cm-1) × (14 - 0) = 27419cm-1υ = 3.40ev$

(b)

Interpretation Introduction

Interpretation:

The ionization energy of hydrogen atom when it is in lower state of transition has to be identified.

(b)

Expert Solution

Explanation of Solution

The energy needed to ionize the atom is obtained by allowing $n2→ ∞$

$υ = 1λ∞ = (109677 cm-1) × (14 - 0) = 27419cm-11cm-1 = 1.2398 ×10-4evυ = 3.40ev$

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Answer 4

Question

Chapter 8, Problem 8.4P

(a)

Interpretation Introduction

Interpretation:

The wavelength of the next line in the given series of lines of spectrum of atomic hydrogen has to be identified.

(a)

Expert Solution

Explanation of Solution

All the lines in the hydrogen spectrum suits the Rydberg formula.

$1λ = RH(1n12 - 1n22)Where λ is the wavelength of the emitted photonR is Rydberg’s constantZ is the atomic number (1 for H, 2 for He)n1 and n2 are the principal quantum numbers for the initial and final states$

Find $n1$ from the value of $λmax$ , which arises from the transition of $n1+ 1 → n1$ .

$1λmaxRH = 1n12 - (1n1 + 1)2 = 2n1 + 1n12 (n1 + 1)2λmaxRH = n12 (n1 + 1)22n1 + 1 = (656.46 × 10-9m) × (109677× 102m-1 ) = 7.20hence n1 = 2, as determined by trial and error substitution. Therefore,the transitions are given as,υ = 1λ = (109677 cm-1) × (14 - 1n22) n2 = 3,4,5,6Next line has n2 = 7υ = 1λ = (109677 cm-1) × (14 - 149) = 397.13nm$

The energy needed to ionize the atom is obtained by allowing $n2→ ∞$

$υ = 1λ∞ = (109677 cm-1) × (14 - 0) = 27419cm-1υ = 3.40ev$

(b)

Interpretation Introduction

Interpretation:

The ionization energy of hydrogen atom when it is in lower state of transition has to be identified.

(b)

Expert Solution

Explanation of Solution

The energy needed to ionize the atom is obtained by allowing $n2→ ∞$

$υ = 1λ∞ = (109677 cm-1) × (14 - 0) = 27419cm-11cm-1 = 1.2398 ×10-4evυ = 3.40ev$

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Answer 5

Chapter 8, Problem 8.4P

(a)

Interpretation Introduction

Interpretation:

The wavelength of the next line in the given series of lines of spectrum of atomic hydrogen has to be identified.

(a)

Expert Solution

Explanation of Solution

All the lines in the hydrogen spectrum suits the Rydberg formula.

$1λ = RH(1n12 - 1n22)Where λ is the wavelength of the emitted photonR is Rydberg’s constantZ is the atomic number (1 for H, 2 for He)n1 and n2 are the principal quantum numbers for the initial and final states$

Find $n1$ from the value of $λmax$ , which arises from the transition of $n1+ 1 → n1$ .

$1λmaxRH = 1n12 - (1n1 + 1)2 = 2n1 + 1n12 (n1 + 1)2λmaxRH = n12 (n1 + 1)22n1 + 1 = (656.46 × 10-9m) × (109677× 102m-1 ) = 7.20hence n1 = 2, as determined by trial and error substitution. Therefore,the transitions are given as,υ = 1λ = (109677 cm-1) × (14 - 1n22) n2 = 3,4,5,6Next line has n2 = 7υ = 1λ = (109677 cm-1) × (14 - 149) = 397.13nm$

The energy needed to ionize the atom is obtained by allowing $n2→ ∞$

$υ = 1λ∞ = (109677 cm-1) × (14 - 0) = 27419cm-1υ = 3.40ev$

(b)

Interpretation Introduction

Interpretation:

The ionization energy of hydrogen atom when it is in lower state of transition has to be identified.

(b)

Expert Solution

Explanation of Solution

The energy needed to ionize the atom is obtained by allowing $n2→ ∞$

$υ = 1λ∞ = (109677 cm-1) × (14 - 0) = 27419cm-11cm-1 = 1.2398 ×10-4evυ = 3.40ev$

Answer 6

Chapter 8, Problem 8.4P

(a)

Interpretation Introduction

Interpretation:

The wavelength of the next line in the given series of lines of spectrum of atomic hydrogen has to be identified.

(a)

Expert Solution

Explanation of Solution

All the lines in the hydrogen spectrum suits the Rydberg formula.

$1λ = RH(1n12 - 1n22)Where λ is the wavelength of the emitted photonR is Rydberg’s constantZ is the atomic number (1 for H, 2 for He)n1 and n2 are the principal quantum numbers for the initial and final states$

Find $n1$ from the value of $λmax$ , which arises from the transition of $n1+ 1 → n1$ .

$1λmaxRH = 1n12 - (1n1 + 1)2 = 2n1 + 1n12 (n1 + 1)2λmaxRH = n12 (n1 + 1)22n1 + 1 = (656.46 × 10-9m) × (109677× 102m-1 ) = 7.20hence n1 = 2, as determined by trial and error substitution. Therefore,the transitions are given as,υ = 1λ = (109677 cm-1) × (14 - 1n22) n2 = 3,4,5,6Next line has n2 = 7υ = 1λ = (109677 cm-1) × (14 - 149) = 397.13nm$

The energy needed to ionize the atom is obtained by allowing $n2→ ∞$

$υ = 1λ∞ = (109677 cm-1) × (14 - 0) = 27419cm-1υ = 3.40ev$

Answer 7

Chapter 8, Problem 8.4P

(a)

Interpretation Introduction

Interpretation:

The wavelength of the next line in the given series of lines of spectrum of atomic hydrogen has to be identified.

Answer 8

(a)

Expert Solution

Explanation of Solution

All the lines in the hydrogen spectrum suits the Rydberg formula.

$1λ = RH(1n12 - 1n22)Where λ is the wavelength of the emitted photonR is Rydberg’s constantZ is the atomic number (1 for H, 2 for He)n1 and n2 are the principal quantum numbers for the initial and final states$

Find $n1$ from the value of $λmax$ , which arises from the transition of $n1+ 1 → n1$ .

$1λmaxRH = 1n12 - (1n1 + 1)2 = 2n1 + 1n12 (n1 + 1)2λmaxRH = n12 (n1 + 1)22n1 + 1 = (656.46 × 10-9m) × (109677× 102m-1 ) = 7.20hence n1 = 2, as determined by trial and error substitution. Therefore,the transitions are given as,υ = 1λ = (109677 cm-1) × (14 - 1n22) n2 = 3,4,5,6Next line has n2 = 7υ = 1λ = (109677 cm-1) × (14 - 149) = 397.13nm$

The energy needed to ionize the atom is obtained by allowing $n2→ ∞$

$υ = 1λ∞ = (109677 cm-1) × (14 - 0) = 27419cm-1υ = 3.40ev$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

Interpretation:

The ionization energy of hydrogen atom when it is in lower state of transition has to be identified.

(b)

Expert Solution

Explanation of Solution

The energy needed to ionize the atom is obtained by allowing $n2→ ∞$

$υ = 1λ∞ = (109677 cm-1) × (14 - 0) = 27419cm-11cm-1 = 1.2398 ×10-4evυ = 3.40ev$

Answer 13

(b)

Interpretation Introduction

Interpretation:

The ionization energy of hydrogen atom when it is in lower state of transition has to be identified.

Answer 14

(b)

Expert Solution

Explanation of Solution

The energy needed to ionize the atom is obtained by allowing $n2→ ∞$

$υ = 1λ∞ = (109677 cm-1) × (14 - 0) = 27419cm-11cm-1 = 1.2398 ×10-4evυ = 3.40ev$

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

Ch. 8 - Prob. 8A.1ST Ch. 8 - Prob. 8A.2ST Ch. 8 - Prob. 8A.3ST Ch. 8 - Prob. 8A.4ST Ch. 8 - Prob. 8A.5ST Ch. 8 - Prob. 8B.2ST Ch. 8 - Prob. 8B.3ST Ch. 8 - Prob. 8B.4ST Ch. 8 - Prob. 8C.1ST Ch. 8 - Prob. 8C.2ST

The wavelength of the next line in the given series of lines of spectrum of atomic hydrogen has to be identified.

Concept explainers

Interpretation:

The wavelength of the next line in the given series of lines of spectrum of atomic hydrogen has to be identified.

Explanation of Solution

Interpretation:

The ionization energy of hydrogen atom when it is in lower state of transition has to be identified.

Explanation of Solution

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Chapter 8 Solutions

The wavelength of the next line in the given series of lines of spectrum of atomic hydrogen has to be identified.

Concept explainers

Interpretation: The wavelength of the next line in the given series of lines of spectrum of atomic hydrogen has to be identified.

Explanation of Solution

Interpretation: The ionization energy of hydrogen atom when it is in lower state of transition has to be identified.

Explanation of Solution

Want to see more full solutions like this?

Chapter 8 Solutions

Interpretation:

The wavelength of the next line in the given series of lines of spectrum of atomic hydrogen has to be identified.

Interpretation:

The ionization energy of hydrogen atom when it is in lower state of transition has to be identified.