   # Suppose hydrogen atoms absorb energy so that electrons are excited to the n = 7 energy level. Electrons then undergo these transitions, among others: (a) n = 7 → n = 1; (b) n = 7 → n = 6; and (c) n = 2 → n = 1. Which of these transitions produces a photon with (i) the smallest energy, (ii) the highest frequency, and (iii) the shortest wavelength? ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 6, Problem 60GQ
Textbook Problem
620 views

## Suppose hydrogen atoms absorb energy so that electrons are excited to the n = 7 energy level. Electrons then undergo these transitions, among others: (a) n = 7 → n = 1; (b) n = 7 → n = 6; and (c) n = 2 → n = 1. Which of these transitions produces a photon with (i) the smallest energy, (ii) the highest frequency, and (iii) the shortest wavelength?

(i)

Interpretation Introduction

Interpretation: The transition having smallest energy (b) highest frequency and (c) shortest wavelength among the given transitions.

Concept introduction:

• Electronic transitions that take place in excited H atom is,
1. a. Lyman series: electronic transitions take place to the n=1 level and it is in ultraviolet region.
1. b. Balmer series: electronic transitions take place from n>2 to the n=2 level and it is in visible region.
2. c. Ritz-Paschen series: electronic transitions take place from n>3 to the n=3 level and it is in infrared region.
3. d. Brackett series: electronic transitions take place from n>4 to the n=4 level.
4. e. Pfund series: electronic transitions take place from n>5 to the n=5 level
• Planck’s equation,

E==hcλwhere, E=energyh=Planck'sconstantν=frequencyλ=wavelengthc=speedoflight

The energy increases as the wavelength of the light decrease. Also the energy increases as the frequency of the light increases.

• As the energy gap between two transition states increases the wavelength of the radiation emitted decreases.

### Explanation of Solution

E==hcλ so the energy increases as the wavelength of the light decrease. So the transition having longest wavelength has lowest energy.

As the energy gap between two transition states increases the wavelength of the radiation emitted decreases

(ii)

Interpretation Introduction

Interpretation: The transition having highest frequency among the given transitions.

Concept introduction:

• Electronic transitions that take place in excited H atom is,
1. a. Lyman series: electronic transitions take place to the n=1 level and it is in ultraviolet region.
1. b. Balmer series: electronic transitions take place from n>2 to the n=2 level and it is in visible region.
2. c. Ritz-Paschen series: electronic transitions take place from n>3 to the n=3 level and it is in infrared region.
3. d. Brackett series: electronic transitions take place from n>4 to the n=4 level.
4. e. Pfund series: electronic transitions take place from n>5 to the n=5 level
• As the energy gap between two transition states increases the frequency of the radiation emitted also increases.

(iii)

Interpretation Introduction

Interpretation: The transition having shortest wavelength among the given transitions.

Concept introduction:

• Electronic transitions that take place in excited H atom is,
1. a. Lyman series: electronic transitions take place to the n=1 level and it is in ultraviolet region.
1. b. Balmer series: electronic transitions take place from n>2 to the n=2 level and it is in visible region.
2. c. Ritz-Paschen series: electronic transitions take place from n>3 to the n=3 level and it is in infrared region.
3. d. Brackett series: electronic transitions take place from n>4 to the n=4 level.
4. e. Pfund series: electronic transitions take place from n>5 to the n=5 level
• As the energy gap between two transition states increases the wavelength of the radiation emitted decreases.

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