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

Question

Want to see more full solutions like this?

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 $n2→ ∞$

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

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

Post Lab Questions. 1) Draw the mechanism of your Diels-Alder cycloaddition. 2) Only one isomer of product is formed in the Diels-Alder cycloaddition. Why? 3) Imagine that you used isoprene as diene - in that case you don't have to worry about assigning endo vs exo. Draw the "endo" and "exo" products of the Diels-Alder reaction between isoprene and maleic anhydride, and explain why the distinction is irrelevant here. 4) This does not hold for other dienes. Draw the exo and endo products of the reaction of cyclohexadiene with maleic anhydride. Make sure you label your answers properly as endo or exo. 100 °C Xylenes ??? 5) Calculate the process mass intensity for your specific reaction (make sure to use your actual amounts of reagent).

Indicate the product(s) A, B C and D that are formed in the reaction: H + NH-NH-CH [A+B] [C+D] hydrazones

How can you prepare a 6 mL solution of 6% H2O2, if we have a bottle of 30% H2O2?

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 $n2→ ∞$

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

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 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 $n2→ ∞$

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

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 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 $n2→ ∞$

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

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 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

Want to see more full solutions like this?

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.