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 7.74P

(a)

Interpretation Introduction

Interpretation:

A general expression for the ionization energy of a one electron species is to be written.

Concept introduction:

Ionization energy is defined as the amount of energy required to remove an electron from an isolated gaseous atom. The energy required to remove an electron from an atom depends on the position of the electron in the atom. The closer the electron is to the nucleus in the atom, the harder it is to pull it out of the atom. As the distance of an electron from the nucleus increases, the magnitude of the forces of attraction between the electron and the nucleus decreases. Thus it becomes easier to remove it from the atom.

The equation to find the difference in the energy between the two levels in hydrogen-like atoms is,

ΔE=2.18×1018 J(1nfinal21ninitial2)Z2 (1)

(a)

Expert Solution
Check Mark

Answer to Problem 7.74P

The general expression for the ionization energy of one mole of a one electron species is ΔE=(1.313014×106)Z2 J/mol.

Explanation of Solution

The ionization energy of an atom is the minimum amount of energy required to completely remove the outermost electron from it. An electron is completely removed from an atom when the value of nfinal is . This is so because the electron is completely free from the influence of the nuclear charge.

Substitute for nfinal and 1 for ninitial in equation (1).

ΔE=(2.18×1018 J)(12112)Z2

For one mole of one electron species, the equation becomes,

ΔE=(2.18×1018 J)(12112)Z2(6.023×10231 mol)=(13.13014×105)Z2 J/mol=(1.313014×106)Z2 J/mol

Conclusion

The general expression for the ionization energy of one mole of a one electron species is ΔE=(1.313014×106)Z2 J/mol.

(b)

Interpretation Introduction

Interpretation:

The ionization energy of B4+ is to be calculated.

Concept introduction:

Ionization energy is defined as the amount of energy required to remove an electron from an isolated gaseous atom. The energy required to remove an electron from an atom depends on the position of the electron in the atom. The closer the electron is to the nucleus in the atom, the harder it is to pull it out of the atom. As the distance of an electron from the nucleus increases, the magnitude of the forces of attraction between the electron and the nucleus decreases. Thus it becomes easier to remove it from the atom.

The general expression for the ionization energy of one mole of a one electron species is ΔE=(1.313014×106)Z2 J/mol (2)

(b)

Expert Solution
Check Mark

Answer to Problem 7.74P

The ionization energy of B4+ is 3.28×107 J/mol.

Explanation of Solution

The symbol B represents the element boron. The Z value signifies the atomic number of an element. The atomic number of boron is 5.

Substitute 5 for Z in eqution (2).

ΔE=(1.313014×106)(5)2 J/mol=(1.313014×106)(25) J/mol=32.82535×106 J/mol=3.28×107 J/mol

Conclusion

The ionization energy of B4+ is 3.28×107 J/mol.

(c)

Interpretation Introduction

Interpretation:

The minimum wavelength required to remove the electron from the n=3 level of He+ is to be determined.

Concept introduction:

Ionization energy is defined as the amount of energy required to remove an electron from an isolated gaseous atom. The energy required to remove an electron from an atom depends on the position of the electron in the atom. The closer the electron is to the nucleus in the atom, the harder it is to pull it out of the atom. As the distance of an electron from the nucleus increases, the magnitude of the forces of attraction between the electron and the nucleus decreases. Thus it becomes easier to remove it from the atom.

The equation that relates to the frequency and wavelength of electromagnetic radiation is as follows:

ν=cλ

Here,

c is the speed of light.

λ is the wavelength.

ν is the frequency of the radiation.

Energy is proportional to the frequency and is expressed by the Plank-Einstein equation as follows:

ΔE=hν

Here,

E is the energy.

h is the Plank’s constant.

ν is the frequency.

The above relation can be modified as follows:

ΔE=h(cλ) (3)

(c)

Expert Solution
Check Mark

Answer to Problem 7.74P

The minimum wavelength required to remove the electron from the n=3 level of He+ is 205 nm.

Explanation of Solution

Substitute for nfinal, 3 for ninitial and 2 for Z in equation (1).

ΔE=(2.18×1018 J)(12132)(22)=0.9688×1018 J=9.688×1019 J

Substitute 6.63×1034 Js for h, 3×108 for c and 9.688×1019 J for ΔE in equation (3).

9.688×1019 J=(6.63×1034 Js)(3×108λ)

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

λ=(6.63×1034 Js)(3×108)9.688×1019 J(1 nm109 m)=2.053×107 m=205×109 m

Conclusion

The minimum wavelength required to remove the electron from the n=3 level of He+ is 205 nm.

(d)

Interpretation Introduction

Interpretation:

The minimum wavelength required to move the electron from n=2 level of Be3+, is to be determined.

Concept introduction:

The equation to find the difference in the energy between the two levels in hydrogen-like atoms is,

ΔE=2.18×1018 J(1nfinal21ninitial2)Z2 (1)

The equation that relates to the frequency and wavelength of electromagnetic radiation is as follows:

ν=cλ

Here,

c is the speed of light.

λ is the wavelength.

ν is the frequency of the radiation.

Energy is proportional to the frequency and is expressed by the Plank-Einstein equation as follows:

ΔE=hν

Here,

E is the energy.

h is the Plank’s constant.

ν is the frequency.

The above relation can be modified as follows:

ΔE=h(cλ) (3)

(d)

Expert Solution
Check Mark

Answer to Problem 7.74P

The minimum wavelength required to move the electron from n=2 level of Be3+, is 22.8 nm.

Explanation of Solution

Substitute for nfinal, 2 for ninitial and 4 for Z in equation (1).

ΔE=(2.18×1018 J)(12122)(42)=(2.18×1018 J)(014)(16)=8.72×1018 J

Substitute 6.63×1034 Js for h, 3×108 for c and 8.72×1018 J for ΔE in equation (3).

8.72×1018 J=(6.63×1034 Js)(3×108λ)

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

λ=(6.63×1034 Js)(3×108)8.72×1018 J(1 nm109 m)=2.28×108 m=22.8×109 m

Conclusion

The minimum wavelength required to move the electron from n=2 level of Be3+, is 22.8 nm.

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Chapter 7 Solutions

MCGRAW: CHEMISTRY THE MOLECULAR NATURE

Ch. 7.3 - Prob. 7.5AFPCh. 7.3 - Prob. 7.5BFPCh. 7.4 - What are the possible l and m1 values for n = 4? Ch. 7.4 - Prob. 7.6BFPCh. 7.4 - Prob. 7.7AFPCh. 7.4 - Prob. 7.7BFPCh. 7.4 - Prob. 7.8AFPCh. 7.4 - Prob. 7.8BFPCh. 7 - Prob. 7.1PCh. 7 - Consider the following types of electromagnetic...Ch. 7 - Prob. 7.3PCh. 7 - In the 17th century, Isaac Newton proposed that...Ch. 7 - Prob. 7.5PCh. 7 - What new idea about light did Einstein use to...Ch. 7 - An AM station broadcasts rock music at “950 on...Ch. 7 - An FM station broadcasts music at 93.5 MHz...Ch. 7 - Prob. 7.9PCh. 7 - An x-ray has a wavelength of 1.3 Å. Calculate the...Ch. 7 - Prob. 7.11PCh. 7 - Prob. 7.12PCh. 7 - Police often monitor traffic with “K-band” radar...Ch. 7 - Covalent bonds in a molecule absorb radiation in...Ch. 7 - Prob. 7.15PCh. 7 - Prob. 7.16PCh. 7 - How is n1 in the Rydberg equation (Equation 7.4)...Ch. 7 - What key assumption of Bohr’s model would a “Solar...Ch. 7 - Prob. 7.19PCh. 7 - Which of these electron transitions correspond to...Ch. 7 - Why couldn’t the Bohr model predict spectra for...Ch. 7 - Prob. 7.22PCh. 7 - Use the Rydberg equation to find the wavelength...Ch. 7 - Prob. 7.24PCh. 7 - Prob. 7.25PCh. 7 - Prob. 7.26PCh. 7 - Prob. 7.27PCh. 7 - Prob. 7.28PCh. 7 - Prob. 7.29PCh. 7 - Prob. 7.30PCh. 7 - Prob. 7.31PCh. 7 - Prob. 7.32PCh. 7 - In addition to continuous radiation, fluorescent...Ch. 7 - Prob. 7.34PCh. 7 - Prob. 7.35PCh. 7 - Prob. 7.36PCh. 7 - Prob. 7.37PCh. 7 - Prob. 7.38PCh. 7 - A 232-lb fullback runs 40 yd at 19.8 ± 0.1...Ch. 7 - Prob. 7.40PCh. 7 - Prob. 7.41PCh. 7 - Prob. 7.42PCh. 7 - Prob. 7.43PCh. 7 - Prob. 7.44PCh. 7 - What physical meaning is attributed to ψ2? Ch. 7 - What does “electron density in a tiny volume of...Ch. 7 - Prob. 7.47PCh. 7 - Prob. 7.48PCh. 7 - How many orbitals in an atom can have each of the...Ch. 7 - Prob. 7.50PCh. 7 - Give all possible ml values for orbitals that have...Ch. 7 - Prob. 7.52PCh. 7 - Prob. 7.53PCh. 7 - Prob. 7.54PCh. 7 - Prob. 7.55PCh. 7 - Prob. 7.56PCh. 7 - Prob. 7.57PCh. 7 - Prob. 7.58PCh. 7 - Prob. 7.59PCh. 7 - Prob. 7.60PCh. 7 - Prob. 7.61PCh. 7 - The quantum-mechanical treatment of the H atom...Ch. 7 - The photoelectric effect is illustrated in a plot...Ch. 7 - Prob. 7.64PCh. 7 - Prob. 7.65PCh. 7 - Prob. 7.66PCh. 7 - Prob. 7.67PCh. 7 - Prob. 7.68PCh. 7 - Prob. 7.69PCh. 7 - Prob. 7.70PCh. 7 - Prob. 7.71PCh. 7 - Prob. 7.72PCh. 7 - Prob. 7.73PCh. 7 - Prob. 7.74PCh. 7 - Use the relative size of the 3s orbital below to...Ch. 7 - Prob. 7.76PCh. 7 - Prob. 7.77PCh. 7 - Enormous numbers of microwave photons are needed...Ch. 7 - Prob. 7.79PCh. 7 - Prob. 7.80PCh. 7 - Prob. 7.81PCh. 7 - Prob. 7.82PCh. 7 - Prob. 7.83PCh. 7 - Prob. 7.84PCh. 7 - For any microscope, the size of the smallest...Ch. 7 - In fireworks, the heat of the reaction of an...Ch. 7 - Prob. 7.87PCh. 7 - Fish-liver oil is a good source of vitamin A,...Ch. 7 - Many calculators use photocells as their energy...Ch. 7 - Prob. 7.90PCh. 7 - Prob. 7.91PCh. 7 - Prob. 7.92PCh. 7 - The flame tests for sodium and potassium are based...Ch. 7 - Prob. 7.94PCh. 7 - Prob. 7.95PCh. 7 - The discharge of phosphate in detergents to the...
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