General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 7, Problem 7.119SP

(a)

Interpretation Introduction

Interpretation:

The electronic transition corresponds to line B and C has to be identified.

(a)

Expert Solution
Check Mark

Explanation of Solution

The given lines are corresponding to n=2.  Line A has longest wavelength or lowest energy transition.  This indicates the transition is 32 transition.  Line B has greater wavelength and lower energy than line C.  Therefore, line B corresponds to 42  and line C corresponds to 52 transition.

(b)

Interpretation Introduction

Interpretation:

From the wavelength of line c, the wavelength of line A and B has to be calculated.

Concept Introduction:

Bohr developed a rule for quantization of energy that could be applicable to the electron of an atom in motion.  By using this, he derived a formula for energy levels of electron in H-atom.

E=-RHn2n=1,2,3,......(Forhydrogenatom)RHisRydbergconstant(2.179×10-18J).Eisenergylevel.nisprincipalquantumnumber.

(b)

Expert Solution
Check Mark

Answer to Problem 7.119SP

The wavelength of line A is 41.1nm

The wavelength of line B is 30.4nm

Explanation of Solution

The energy of line C is calculated using its wavelength as follows,

E=hcλ=(6.63×10-34J.s)(3.00×108m/s)(27.1×10-9m) =7.34×10-18J

The atom in which electronic transition of 52 occurs is identified as

ΔE==Ef-EiΔE=RHZ2(1ni2-1nf2)-7.34×10-18J=RHZ2(1ni2-1nf2)-7.34×10-18J=(2.18×10-18J)Z2(152-122)-7.34×10-18J=Z2(2.18×10-18J52-2.18×10-18J22)-7.34×10-18J=Z2(8.72×10-18J-5.45×10-17J100)-7.34×10-18J=(-4.58×10-19)Z2Z2= 16.0Z=4

The energy change and wavelength for transitions of 32 and 42 are calculated as

ΔE=RHZ2(1ni2-1nf2)=(2.18×10-18J)(4)2(132-122)=(3.488×10-17J)(132-122)    =(3.488×10-17J32-3.488×10-17J22)    =(1.395×10-16J- 3.139×10-1636)=-4.84×10-18J

Negative sign is neglected while calculating lambda

λ=hcΔE=(6.63×10-34J.s)(3.00×108m/s)4.84×10-18Jλ=4.11×10-8m=41.1nm1m=109nm

ΔE=RHZ2(1ni2-1nf2)=(2.18×10-18J)(4)2(142-122)=3.488×10-17(316)=-6.54×10-18J

Negative sign is neglected while calculating lambda

λ=hcΔE=(6.63×10-34J.s)(3.00×108m/s)6.54×10-18Jλ=3.04×10-8m=30.4nm(1m=109nm)

(c)

Interpretation Introduction

Interpretation:

The energy required to eject an electron from n=4 has to be calculated.

Concept Introduction:

Bohr developed a rule for quantization of energy that could be applicable to the electron of an atom in motion.  By using this, he derived a formula for energy levels of electron in H-atom.

E=-RHn2n=1,2,3,......(Forhydrogenatom)RHisRydbergconstant(2.179×10-18J).Eisenergylevel.nisprincipalquantumnumber.

(c)

Expert Solution
Check Mark

Answer to Problem 7.119SP

The energy required to eject an electron from n=4 is 2.18×10-18J

Explanation of Solution

The initial state is n=4 and final state is infinity.  The energy change for transitions of 4 is calculated as

ΔE=RHZ2(1ni2-1nf2)=(2.18×10-18J)(4)2(142-12)=(2.18×10-18J)(4)2(116-0)=(2.18×10-18J)(16)(116)=2.18×10-18J

(d)

Interpretation Introduction

Interpretation:

The physical significance of continuum has to be explained.

(d)

Expert Solution
Check Mark

Explanation of Solution

The energy levels are closely packed when the n values become larger and it leads to continuum of lines.  Electrons have been removed from atom when the continuum starts.  Therefore, there will be no quantized energy levels associated with electron.

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Chapter 7 Solutions

General Chemistry

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