Physical Chemistry
Physical Chemistry
2nd Edition
ISBN: 9781133958437
Author: Ball, David W. (david Warren), BAER, Tomas
Publisher: Wadsworth Cengage Learning,
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Chapter 11, Problem 11.67E

To four significant figures, the first four lines in the Balmer series in the hydrogen atom ( n 2 = 2 ) spectrum appear at 656.5 , 486.3 , 434.2 , and 410.3 nm . (a) From these numbers, calculate an average value of R H , Rydberg constant. (b) At what wavelengths would similar transitions appear for He + ?

Expert Solution
Check Mark
Interpretation Introduction

(a)

Interpretation:

The average value of RH, the Rydberg constant is to be calculated.

Concept introduction:

Rydberg equation is used to represent the wavenumber or wavelength or the energy difference of the lines present in the atomic spectrum of an element. The Rydberg equation for the hydrogen atom is represented as,

ΔE=RH(1n221n12)

Where,

RH represents Rydberg constant with a value for hydrogen 109737.31cm1.

n1 and n2 represents the energy levels.

ΔE is the energy difference.

Answer to Problem 11.67E

The average value of RH, the Rydberg constant is 2.18×1018J.

Explanation of Solution

For Balmer series in the hydrogen atom n2=2. For the first four lines, that is, for n1=3,4,5 and 6 the spectrum appears at 656.5, 486.3, 434.2, and 410.3nm respectively. The expression of wavelength using Rydberg constant is,

ΔE=hcλ=RH(1n221n12)

Where,

RH represents Rydberg constant with a value for hydrogen 109737.31cm1.

n1 and n2 represents the energy levels.

λ is the wavelength for the transition from one from one energy level to another.

h is the Planck’s constant.

ΔE is the energy difference.

Substitute the values of wavelength for n1=3, the Rydberg constant in the given formula.

RH=(6.626×1034Js)(3×108ms1)(656.5×109m)(122132)RH=(6.626×1034Js)(3×108ms1)(656.5×109m)(1419)RH=2.18×1018J

Substitute the values of wavelength for n1=4, the Rydberg constant in the given formula.

RH=(6.626×1034Js)(3×108ms1)(486.3×109m)(122142)RH=(6.626×1034Js)(3×108ms1)(486.3×109m)(14116)RH=2.18×1018J

Substitute the values of wavelength for n1=5, the Rydberg constant in the given formula.

RH=(6.626×1034Js)(3×108ms1)(434.2×109m)(122152)RH=(6.626×1034Js)(3×108ms1)(434.2×109m)(14125)RH=2.18×1018J

Substitute the values of wavelength for n1=6, the Rydberg constant in the given formula.

RH=(6.626×1034Js)(3×108ms1)(410.3×109m)(122162)RH=(6.626×1034Js)(3×108ms1)(410.3×109m)(14136)RH=2.18×1018J

Since all the values of RH are the same, the average value of RH, Rydberg constant is 2.18×1018J.

Conclusion

The average value of RH, the Rydberg constant is 2.18×1018J.

Expert Solution
Check Mark
Interpretation Introduction

(b)

Interpretation:

The wavelengths for which similar transitions occur for He+ are to be calculated.

Concept introduction:

Rydberg equation is used to represent the wavenumber or wavelength or the energy difference of the lines present in the atomic spectrum of an element. The Rydberg equation for the hydrogen atom is represented as,

ΔE=RH(1n221n12)

Where,

RH represents Rydberg constant with a value for hydrogen 109737.31cm1.

n1 and n2 represents the energy levels.

ΔE is the energy difference.

Answer to Problem 11.67E

The wavelength of transition for helium cation for n1=3,4,5 and 6 the spectrum appears at 164.125nm, 121.575nm, 108.550nm, and 102.575nm respectively.

Explanation of Solution

For Balmer series in the hydrogen atom n2=2. For the first four lines, that is, for n1=3,4,5 and 6 the spectrum appears at 656.5, 486.3, 434.2, and 410.3nm.

For hydrogen atom the energy difference is calculated by,ΔEH=e4μ802h2(1n221n12)

Where,

μ is the reduced mass.

e is the charge on electron.

0permittivity of free space.

h is the Planck’s constant.

n1 and n2 represents the energy levels.

The energy difference using Rydberg constant for atoms other than hydrogen is calculated by,ΔE=Z2e4μ802h2(1n221n12)

Where,

Z is the atomic number.

μ is the reduced mass.

e is the charge on electron.

0 is the permittivity of free space.

h is the Planck’s constant.

n1 and n2 represents the energy levels.

For helium atom, Z=2.

Thus, relation between the energy difference of hydrogen and helium cation is,

ΔEHe=22ΔEHΔEHe=4ΔEH

The relation between the wavelengths of transition of hydrogen and helium cation is,

hcλHe=4hcλH

λHe=λH4 …(1)

Substitute the value of wavelength for n1=3, in equation (1).

λHe=656.5nm4λHe=164.125nm

Substitute the value of wavelength for n1=4, in equation (1).

λHe=486.3nm4λHe=121.575nm

Substitute the value of wavelength for n1=5, in equation (1).

λHe=434.2nm4λHe=108.55nm

Substitute the value of wavelength for n1=5, in equation (1).

λHe=410.3nm4λHe=102.575nm

Thus, the wavelength of transition for helium cation for n1=3,4,5 and 6 the spectrum appears at 164.125nm, 121.575nm, 108.550nm, and 102.575nm respectively.

Conclusion

The wavelength of transition for helium cation for n1=3,4,5 and 6 the spectrum appears at 164.125nm, 121.575nm, 108.550nm, and 102.575nm respectively.

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Chapter 11 Solutions

Physical Chemistry

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