Chemistry: The Molecular Nature of Matter and Change
Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781259631757
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
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Chapter 7, Problem 7.80P

(a)

Interpretation Introduction

Interpretation:

Whether the n=1 range of wavelengths series for the hydrogen atom overlaps with the range n=2 wavelength series is 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 wavelength from nm to m is,

1nm=1×109 m

(a)

Expert Solution
Check Mark

Answer to Problem 7.80P

The n1=1 range of wavelengths series for the hydrogen atom does not overlap with the range n1=2 wavelength series

Explanation of Solution

The overlap between n1=1 and n1=2 series is checked by comparing the longest wavelength for n1=1 and the shortest wavelength for n1=2.

The longest wavelength in the n1=1 series has n2=2.

Substitute 1.096776×107 m1 for R,1 for n1 and 2 for n2 in equation (1).

1λ=(1.096776×107 m1)(112122)=(1.096776×107 m1)(1114)=0.822582×107 m1λ=1.215×107 m

The shortest wavelength in the n1=2 series has n2=.

Substitute 1.096776×107 m1 for R,2 for n1 and for n2 in equation (1).

1λ=(1.096776×107 m1)(1221)=(1.096776×107 m1)(140)=0.274194×107 m1λ=3.647×107 m

The longest wavelength for n1=1 series is shorter than the shortest wavelength for n1=2 series. Hence, the overlap is not possible.

Conclusion

The n=1 range of wavelengths series for the hydrogen atom does not overlap with the range n=2 wavelength series

(b)

Interpretation Introduction

Interpretation:

Whether the n=3 range of wavelengths series for the hydrogen atom overlaps with the range n=4 wavelength series is 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.

(b)

Expert Solution
Check Mark

Answer to Problem 7.80P

The n=4 range of wavelengths series for the hydrogen atom does overlap with the range n=3 wavelength series

Explanation of Solution

The overlap between n1=3 and n1=4 series is checked by comparing the longest wavelength for n1=3 and the shortest wavelength for n1=4.

The longest wavelength in the n1=3 series has n2=4.

Substitute 1.096776×107 m1 for R, 3 for n1 and 4 for n2 in equation (1).

1λ=(1.096776×107 m1)(132142)=(1.096776×107 m1)(19116)=0.0533155×107 m1λ=1.875627×106 m

The shortest wavelength in the n1=4 series has n2=.

Substitute 1.096776×107 m1 for R,4 for n1 and for n2 in equation (1).

1λ=(1.096776×107 m1)(1421)=(1.096776×107 m1)(1160)=0.0685485×107 m1λ=1.458821×107 m

The longest wavelength for n1=3 series is longer than the shortest wavelength for n1=4 series. Hence, the overlap is possible.

Conclusion

The n=4 range of wavelengths series for the hydrogen atom does overlap with the range n=3 wavelength series

(c)

Interpretation Introduction

Interpretation:

The number of lines in the n1=4 series that lie in the range of the n1=5 series is 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.

(c)

Expert Solution
Check Mark

Answer to Problem 7.80P

The number of lines in the n1=4 series that lie in the range of the n1=5 series is 2.

Explanation of Solution

The shortest wavelength in the n1=5 series has n2=.

Substitute 1.096776×107 m1 for R,5 for n1 and for n2 in equation (1).

1λ=(1.096776×107 m1)(1521)=(1.096776×107 m1)(1250)=0.04387104×107 m1λ=2.2794×107 m

In the n1=4 series, a few longest wavelengths are calculated to determine whether overlap is possible with the shortest wavelength for n1=5.

The longest wavelength in the n1=4 series has n2=5.

Substitute 1.096776×107 m1 for R, 4 for n1 and 5 for n2 in equation (1).

1λ=(1.096776×107 m1)(142152)=(1.096776×107 m1)(116125)=0.02467746×107 m1λ=4.0522×106 m

For n1=4 and n2=6,

Substitute 1.096776×107 m1 for R, 4 for n1 and 6 for n2 in equation (1).

1λ=(1.096776×107 m1)(142162)=(1.096776×107 m1)(116136)=0.0381678×107 m1λ=2.6200095×106 m

For n1=4 and n2=7,

Substitute 1.096776×107 m1 for R, 4 for n1 and 7 for n2 in equation (1).

1λ=(1.096776×107 m1)(142172)=(1.096776×107 m1)(116149)=0.0461742×107 m1λ=2.165711×106 m

The first two lines with n2 values of 5 and 6 overlap with the shortest wavelength in the n1=5 series.

Conclusion

The number of lines in the n1=4 series that lie in the range of the n1=5 series is 2.

(d)

Interpretation Introduction

Interpretation:

The implication about the hydrogen atom line spectrum made by the overlap at longer wavelengths is 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.

(d)

Expert Solution
Check Mark

Answer to Problem 7.80P

The implication about the hydrogen atom line spectrum made by the overlap at longer wavelengths is that the hydrogen spectrum becomes more complex at longer wavelengths.

Explanation of Solution

The longer wavelengths of a series have more overlapping with the short wavelengths of the successive series. The overlapping of the lines leads to the formation of a continuous spectrum in the form of band. This makes it difficult for the analyst to interpret the needed information. Hence, the overlapping of the lines leads to the complexity of the hydrogen spectrum.

Conclusion

The implication about the hydrogen atom line spectrum made by the overlap at longer wavelengths is that the hydrogen spectrum becomes more complex at longer wavelengths.

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Chapter 7 Solutions

Chemistry: The Molecular Nature of Matter and Change

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