   # The energy emitted when an electron moves from a higher energy state to a lower energy state in any atom can be observed as electromagnetic radiation. (a) Which involves the emission of less energy in the H atom, an electron moving from n = 4 to n = 2 or an electron moving from n = 3 to n = 2? (b) Which involves the emission of more energy in the H atom, an electron moving from n = 4 to n = 1 or an electron moving from n = 5 to n = 2? Explain fully. ### 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 19PS
Textbook Problem
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## The energy emitted when an electron moves from a higher energy state to a lower energy state in any atom can be observed as electromagnetic radiation. (a) Which involves the emission of less energy in the H atom, an electron moving from n = 4 to n = 2 or an electron moving from n = 3 to n = 2? (b) Which involves the emission of more energy in the H atom, an electron moving from n = 4 to n = 1 or an electron moving from n = 5 to n = 2? Explain fully.

(a)

Interpretation Introduction

Interpretation: The transition having lowest energy among n=4 to the n=2 level and n=3 to the n=2 level has to be determined.

Concept introduction:

• Electronic transitions that take place in excited H atom is,

Lyman series: electronic transitions take place to the n=1 level and it is in ultraviolet region.

Balmer series: electronic transitions take place from n>2 to the n=2 level and it is in visible region.

Ritz-Paschen series: electronic transitions take place from n>3 to the n=3 level and it is in infrared region.

Brackett series: electronic transitions take place from n>4 to the n=4 level.

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 frequency of the radiation emitted also increases (E=hν)

### 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 frequency of the radiation emitted also increases (E&#

(b)

Interpretation Introduction

Interpretation: The transition having highest energy among n=4 to the n=1 level and from n=5 to the n=2 level has to be determined.

Concept introduction:

• Electronic transitions that take place in excited H atom is,

Lyman series: electronic transitions take place to the n=1 level and it is in ultraviolet region.

Balmer series: electronic transitions take place from n>2 to the n=2 level and it is in visible region.

Ritz-Paschen series: electronic transitions take place from n>3 to the n=3 level and it is in infrared region.

Brackett series: electronic transitions take place from n>4 to the n=4 level.

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 frequency of the radiation emitted also increases (E=hν).

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