Chapter 6.3, Problem 6.4CYU

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# The Lyman series of spectral lines for the H atom, in the ultraviolet region, arises from transitions from higher levels to n = 1. Calculate the frequency and wavelength of the least energetic line in this series.

Interpretation Introduction

Interpretation:

The frequency and wavelength of least energetic line in Lyman series of the excited H atom has to be calculated.

Concept introduction:

• The energy difference between two states is calculated by using following formula,

EnergybetweenthestatesΔE=EfinalEinitial=Rhc(1nfinal21ninitial2)where,R=Rydbergconstanth=Planck'sconstantc=speedoflightn=Principalquantumnumber

• Planck’s equation,

E=where, E=energyh=Planck'sconstantν=frequency

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

• The frequency of the light is inversely proportional to its wavelength.

ν=cλwhere, c=speedoflightν=frequencyλ=wavelength

Electronic transitions that take place in excited H atom is,

1. 1. Lyman series: electronic transitions take place to the n=1 level and it is in ultraviolet region.
2. 2. Balmer series: electronic transitions take place from n>2 to the n=2 level and it is in visible region.
3. 3. Ritz-Paschen series: electronic transitions take place from n>3 to the n=3 level and it is in infrared region.
4. 4. Brackett series: electronic transitions take place from n>4 to the n=4 level.
5. 5. Pfund series: electronic transitions take place from n>5 to the n=5 level
Explanation

The frequency and wavelength of least energetic line in Lyman series of the excited H atom is calculated.

Given,

The least energetic line in Lyman series of the excited H atom, if the transition of electron is from nâ€‰=â€‰2 to nâ€‰=â€‰1

Â Â Râ€‰=â€‰1.097â€‰Ã—â€‰107â€‰mâ€‰âˆ’1hâ€‰=â€‰6.626â€‰Ã—â€‰10â€‰-34â€‰J.scâ€‰=â€‰2.998â€‰Ã—â€‰10â€‰8â€‰m/sninitialâ€‰=â€‰2nfinalâ€‰=â€‰1

The energy difference between states while emitting photons is calculated by the equation,

Energyâ€‰betweenâ€‰theâ€‰statesâ€‰â€‰Î”Eâ€‰=â€‰Efinalâ€‰âˆ’â€‰Einitial=â€‰âˆ’Rhc(1nfinal2â€‰âˆ’â€‰1ninitial2)

Substituting the values

â€‰Î”Eâ€‰=â€‰â€‰âˆ’â€‰(1.097â€‰Ã—â€‰107â€‰mâ€‰âˆ’1)â€‰Ã—â€‰(6.626â€‰Ã—â€‰10â€‰âˆ’34â€‰J.s)â€‰Ã—â€‰(2.998â€‰Ã—â€‰10â€‰8â€‰m/s)â€‰(112â€‰âˆ’â€‰122)=â€‰âˆ’1.6343â€‰Ã—â€‰10â€‰âˆ’18â€‰J/atom

• The frequency of least energetic line in Lyman series of the excited H atom is calculated,

According to Planckâ€™s equation

â€‚Â Ephotonâ€‰=â€‰hÎ½

Rearranging the equation,

Â Â Â Â Î½â€‰=â€‰Ephotonh=â€‰1

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