MCGRAW: CHEMISTRY THE MOLECULAR NATURE
MCGRAW: CHEMISTRY THE MOLECULAR NATURE
8th Edition
ISBN: 9781264330430
Author: VALUE EDITION
Publisher: MCG
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Chapter 7, Problem 7.68P
Interpretation Introduction

Interpretation:

The ninitial values for each of the five lines in the hydrogen atom spectrum are to be determined.

Concept introduction:

An atom of hydrogen contains one electron. But the spectrum of hydrogen consists of a large number of lines. This is so because a sample of hydrogen contains a very large number of atoms. When energy is supplied to a sample of gaseous atoms of hydrogen, different atoms absorb different amounts of energy. Therefore, the electrons in different atoms jump to different energy levels. Upon losing the energies gained initially, the electrons jump back to lower energy levels and release radiations of different wavelengths.

The equation used to predict the position and wavelength of any line in a given series is called the Rydberg’s equation.

Rydberg’s equation is as follows:

1λ=R(1n121n22) (1)

Here,

λ  is the wavelength of the line.

n1 and  n2 are positive integers, with n2>n1.

R is the Rydberg’s constant.

The conversion factor to convert the wavelength from Ao to m is as follows:

1Ao=1×1010 m

Expert Solution & Answer
Check Mark

Answer to Problem 7.68P

For the lines of wavelengths 1212.7 Ao, 4340.5 Ao, 4861.3 Ao, 6562.8 Ao and 10938 Ao, the values of ninitial are 2, 5, 4, 3 and 6 respectively.

Explanation of Solution

The radiation of wavelength 1212.7 Ao falls within the ultraviolet region of the electromagnetic spectrum. The ultraviolet series of emission lines are obtained when the value of nfinal in the Rydberg’s constant is 1.

Substitute 1212.7 Ao for λ , 1.096776×107 m1 for R and 1 for n1 in equation (1).

11212.7 Ao(1010 m1 Ao)=(1.096776×107 m1)(1121n22)11.2127×107 m=(1.096776×107 m1)(1121n22)0.7518456=(11n22)

Rearrange the above equation and calculate the value for n2 as follows:

(1n22)=10.7518456(1n22)=0.2481544n22=4.029749n2=2

Thus, for the line of wavelength 1212.7 Ao, ninitial is 2.

The radiation of wavelengths 4340.5 Ao, 4861.3 Ao and 6562.8 Ao fall within the visible region of the electromagnetic spectrum. The visible series of emission lines are obtained when the value of nfinal in the Rydberg’s constant is 2.

Substitute 4340.5 Ao for λ , 1.096776×107 m1 for R and 2 for n1 in equation (1).

14340.5 Ao(1010 m1 Ao)=(1.096776×107 m1)(1221n22)14.3405×107 m=(1.096776×107 m1)(141n22)0.2100594=(141n22)

Rearrange the above equation and calculate the value for n2 as follows:

(1n22)=140.2100594(1n22)=0.0399406n22=25.037180n2=5

Thus, for the line of wavelength 4340.5 Ao, ninitial is 5.

Substitute 4861.3 Ao for λ , 1.096776×107 m1 for R and 2 for n1 in equation (1).

14861.3 Ao(1010 m1 Ao)=(1.096776×107 m1)(1221n22)14.8613×107 m=(1.096776×107 m1)(141n22)0.1875554=(141n22)

Rearrange the above equation and calculate the value for n2 as follows:

(1n22)=140.1875554(1n22)=0.0624446n22=16.0141949n2=4

Thus, for the line of wavelength 4861.3 Ao, ninitial is 4.

Substitute 6562.8 Ao for λ , 1.096776×107 m1 for R and 2 for n1 in equation (1).

16562.8 Ao(1010 m1 Ao)=(1.096776×107 m1)(1221n22)16.5628×107 m=(1.096776×107 m1)(141n22)0.1523739=(141n22)

Rearrange the above equation and calculate the value for n2 as follows:

(1n22)=140.1523739(1n22)=0.0976261n22=10.2430n2=3

Thus, for the line of wavelength 6562.8 Ao, ninitial is 3.

The radiation of wavelength 10938 Ao falls within the infrared region of the electromagnetic spectrum. The infrared series of emission lines are obtained when the value of nfinal in the Rydberg’s constant is 3.

Substitute 10938 Ao for λ , 1.096776×107 m1 for R and 3 for n1 in equation (1).

110938 Ao(1010 m1 Ao)=(1.096776×107 m1)(1321n22)110.938×107 m=(1.096776×107 m1)(191n22)0.0833573=(191n22)

Rearrange the above equation and calculate the value for n2 as follows:

(1n22)=190.0833573(1n22)=0.0277538n22=36.031102n2=6

Thus, for the line of wavelength 10938 Ao, ninitial is 6.

Conclusion

For the lines of wavelengths 1212.7 Ao, 4340.5 Ao, 4861.3 Ao, 6562.8 Ao and 10938 Ao, the values of ninitial are 2, 5, 4, 3 and 6 respectively.

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Chapter 7 Solutions

MCGRAW: CHEMISTRY THE MOLECULAR NATURE

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