GENERAL CHEMISTRY ACHIEVE ACCESS W/BOOK
GENERAL CHEMISTRY ACHIEVE ACCESS W/BOOK
4th Edition
ISBN: 9781319405212
Author: McQuarrie
Publisher: MAC HIGHER
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Chapter 4, Problem 4.66P
Interpretation Introduction

Interpretation:

Frequency of the first five lines in Lyman series has to be calculated.

Expert Solution & Answer
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Explanation of Solution

Frequency of First line:

Energy of photon emitted on relaxation from n=2 state to n=1 state in first line of Lyman series and is calculated as shown below.

    Ephoton=(2.1799×1018J)(1nf21ni2)=(2.1799×1018J)(112122)=(2.1799×1018J)(1114)=1.6349×1018J

Frequency of photon is calculated as shown below.

    E=hvv=Eh=1.6349×1018J6.626×1034Js=0.246740114699668×1016s-1=2.467×1015s1

Therefore, the frequency of photon is calculated as 2.467×1015s1.

Wavelength of photon is calculated as shown below;

    λ=cv=2.9979×108ms12.467×1015s1=1.215×107m=121.5×107cm

Therefore, wavelength of photon is 121.5×107cm.

Frequency of Second line:

Energy of photon emitted on relaxation from n=3 state to n=1 state in second line of Lyman series and is calculated as shown below.

    Ephoton=(2.1799×1018J)(1nf21ni2)=(2.1799×1018J)(112132)=(2.1799×1018J)(1119)=1.9377×1018J

Frequency of photon is calculated as shown below.

    E=hvv=Eh=1.9377×1018J6.626×1034Js=0.2924388771506188×1016s-1=2.924×1015s1

Therefore, the frequency of photon is calculated as 2.924×1015s1.

Wavelength of photon is calculated as shown below;

    λ=cv=2.9979×108ms12.924×1015s1=1.025×107m=102.5×107cm

Therefore, wavelength of photon is 102.5×107cm.

Frequency of Third line:

Energy of photon emitted on relaxation from n=4 state to n=1 state in third line of Lyman series and is calculated as shown below.

    Ephoton=(2.1799×1018J)(1nf21ni2)=(2.1799×1018J)(112142)=(2.1799×1018J)(11116)=2.0436×1018J

Frequency of photon is calculated as shown below.

    E=hvv=Eh=2.0436×1018J6.626×1034Js=0.3084213703591911×1016s-1=3.08×1015s1

Therefore, the frequency of photon is calculated as 3.08×1015s1.

Wavelength of photon is calculated as shown below;

    λ=cv=2.9979×108ms13.08×1015s1=0.9733×107m=97.33×107cm

Therefore, wavelength of photon is 97.33×107cm.

Frequency of Fourth line:

Energy of photon emitted on relaxation from n=5 state to n=1 state in fourth line of Lyman series and is calculated as shown below.

    Ephoton=(2.1799×1018J)(1nf21ni2)=(2.1799×1018J)(112152)=(2.1799×1018J)(11125)=2.0927×1018J

Frequency of photon is calculated as shown below.

    E=hvv=Eh=2.0927×1018J6.626×1034Js=0.3158315725928162×1016s-1=3.16×1015s1

Therefore, the frequency of photon is calculated as 3.16×1015s1.

Wavelength of photon is calculated as shown below;

    λ=cv=2.9979×108ms13.16×1015s1=0.9487×107m=94.87×107cm

Therefore, wavelength of photon is 94.87×107cm.

Frequency of Fifth line:

Energy of photon emitted on relaxation from n=6 state to n=1 state in fifth line of Lyman series and is calculated as shown below.

    Ephoton=(2.1799×1018J)(1nf21ni2)=(2.1799×1018J)(112162)=(2.1799×1018J)(11136)=2.1193×1018J

Frequency of photon is calculated as shown below.

    E=hvv=Eh=2.1193×1018J6.626×1034Js=0.3198460609719288×1016s-1=3.19×1015s1

Therefore, the frequency of photon is calculated as 3.19×1015s1.

Wavelength of photon is calculated as shown below;

    λ=cv=2.9979×108ms13.19×1015s1=0.9397×107m=93.97×107cm

Therefore, wavelength of photon is 93.97×107cm.

Summary of the wavelength of first five lines in Lyman series is given below.

ni23456
Wavelength/×10-7 cm121.5102.597.3394.8793.97

From the above table, in order to plot graph, the wavenumber and 1/ni2 is calculated and summarized as shown below;

1/ni20.250.110.060.040.03
Wavenumber (cm-1)8230497560102743105407106416

Graph is plotted considering wavenumber in y axis and 1/ni2 in x axis as shown;

GENERAL CHEMISTRY ACHIEVE ACCESS W/BOOK, Chapter 4, Problem 4.66P

From the graph that is plotted as shown, the slope of the graph is 109339cm1 and this value is approximately equal to the Rydberg constant.

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