The value ψ 2 , ψ 2 and 4 π r 2 ψ 2 for different value of r are to be calculated. Also, the graph between ψ versus radius, ψ 2 versus radius and 4πr 2 ψ 2 versus radius is to be drawn. Concept introduction: Quantum mechanics helps to understand the wave nature of particles. According to de Broglie, particles like electrons exhibit wave-like properties along with their particle properties. The Schrodinger wave equation is represented as follows: H ∧ ψ = E ψ Here, E is the energy of the atom. H ∧ is the Hamiltonian operator. ψ is the wave function. In the Schrodinger wave equation, ψ represents the wave function. The wave function represents the wave associated with a particle. It has no physical significance. However, the square of the wave function ( ψ 2 ) gives the probability of finding an electron at a particular point and time. Thus it becomes possible to determine the region in an atom around the nucleus where there is a high possibility of finding an electron with a certain amount of energy. This region is called an atomic orbital. The symbol ψ 2 denotes the probability density. It gives the probability of finding an electron in an extremely small region in the atom The expression to calculate the wave number of 1 s the orbital is as follows: ψ = 1 π ( 1 a o ) 3 2 e − r a o (1) Here, ψ is the wave function. r is the distance from the nucleus.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 7, Problem 7.79P

Interpretation Introduction

Interpretation:

The value $ψ2$ , $ψ2$ and $4πr2ψ2$ for different value of $r$ are to be calculated. Also, the graph between $ψ$ versus radius, $ψ2$ versus radius and $4πr2ψ2$ versus radius is to be drawn.

Concept introduction:

Quantum mechanics helps to understand the wave nature of particles. According to de Broglie, particles like electrons exhibit wave-like properties along with their particle properties.

The Schrodinger wave equation is represented as follows:

$H∧ψ=Eψ$

Here,

$E$ is the energy of the atom.

$H∧$ is the Hamiltonian operator.

$ψ$ is the wave function.

In the Schrodinger wave equation, $ψ$ represents the wave function. The wave function represents the wave associated with a particle. It has no physical significance.

However, the square of the wave function $(ψ2)$ gives the probability of finding an electron at a particular point and time. Thus it becomes possible to determine the region in an atom around the nucleus where there is a high possibility of finding an electron with a certain amount of energy. This region is called an atomic orbital.

The symbol $ψ2$ denotes the probability density. It gives the probability of finding an electron in an extremely small region in the atom

The expression to calculate the wave number of $1s$ the orbital is as follows:

$ψ=1π(1ao)32e−rao$ (1)

Here,

$ψ$ is the wave function.

$r$ is the distance from the nucleus.

Expert Solution & Answer

Answer to Problem 7.79P

The value of $ψ2$ , $ψ2$ and $4πr2ψ2$ for different value of $r$ are as follows:

$r(pm)ψψ24πr2ψ201.47×10−32.15×10−60500.570×10−30.325×10−61.02×10−21000.221×10−30.0491×10−60.0616×10−22000.0335×10−30.00112×10−60.0563×10−2$

The graph between $ψ$ versus radius, $ψ2$ versus radius and $4πr2ψ2$ versus r is as follows:

Connect 2-Year Access Card for Chemistry: The Molecular Nature of Matter and Change, Chapter 7, Problem 7.79P , additional homework tip 1

Explanation of Solution

Substitute $52.92 pm$ for $ao$ , $0$ for $r$ in the equation (1).

$ψ=1π(152.92 pm)32e−052.92 pm=1.465532×10−3=1.47×10−3$

The expression to calculate the probability density of electrons is as follows:

$Probability density of electron=ψ2$ (2)

Substitute $1.465532×10−3$ for $ψ$ in the equation (2).

$Probability density of electron=(1.465532×10−3)2=2.15×10−6 pm−3$

The expression to calculate the mass of an electron is as follows:

$Mass of electron=4πr2ψ2$ (3)

Substitute $2.15×10−6 pm−3$ for $ψ2$ and $0$ for $r$ in the equation (3).

$Mass of electron=4π(0)2(2.15×10−6 pm−3)=0$

Substitute $52.92 pm$ for $ao$ , $50 pm$ for $r$ in the equation (1).

$ψ=1π(152.92 pm)32e−50 pm52.92 pm=1.465532(e−50 pm52.92 pm)=5.69724×10−4$

Substitute $5.69724×10−4$ for $ψ$ in the equation (2).

$Probability density of electron=(5.69724×10−4)2=3.24585×10−7 pm−3$

Substitute $3.24585×10−7$ for $ψ2$ and $50 pm$ for $r$ in the equation (3).

$Mass of electron=4π(50 pm)2(3.24585×10−7)=1.0197×10−2 pm−1$

Substitute $52.92 pm$ for $ao$ , $100 pm$ for $r$ in the equation (1).

$ψ=1π(152.92 pm)32e−100 pm52.92 pm=1.465532(e−50 pm52.92 pm)=0.221×10−3$

Substitute $0.221×10−3$ for $ψ$ in the equation (2).

$Probability density of electron=(0.221×10−3)2=0.0491×10−6 pm−3$

Substitute $0.0491×10−6 pm−3$ for $ψ2$ and $50 pm$ for $r$ in the equation (3).

$Mass of electron=4π(50 pm)2(0.0491×10−6 pm−3)=0.616×10−2 pm−1$

Substitute $52.92 pm$ for $ao$ , $200 pm$ for $r$ in the equation (1).

$ψ=1π(152.92 pm)32e−200 pm52.92 pm=1.465532(e−50 pm52.92 pm)=0.0335×10−3$

Substitute $0.0335×10−3$ for $ψ$ in the equation (2).

$Probability density of electron=(0.0335×10−3)2=0.00112×10−6 pm−3$

Substitute $0.00112×10−6 pm−3$ for $ψ2$ and $200 pm$ for $r$ in the equation (3).

$Mass of electron=4π(50 pm)2(0.00112×10−6 pm−3)=0.0563×10−2 pm−1.$

The graph between $ψ$ versus radius, $ψ2$ versus radius and $4πr2ψ2$ versus r is as follows:

Connect 2-Year Access Card for Chemistry: The Molecular Nature of Matter and Change, Chapter 7, Problem 7.79P , additional homework tip 2

Conclusion

The graph between $ψ2$ versus radius plot has zero nodes.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

What is the [OH⁻] of a 1.80 M solution of pyridine (C₅H₅N, Kb = 1.70 × 10⁻⁹)?

What is the percent ionization in a 0.260 M solution of formic acid (HCOOH) (Ka = 1.78 × 10⁻⁴)?

Determine the pH of solution of HC3H5O2 By constructing an ICE table writing the equilibrium constant expression, and using this information to determine the pH. The Ka of HC3H5O2 is 1.3 x 10-5

Answer 2

Question

Chapter 7, Problem 7.79P

Interpretation Introduction

Interpretation:

The value $ψ2$ , $ψ2$ and $4πr2ψ2$ for different value of $r$ are to be calculated. Also, the graph between $ψ$ versus radius, $ψ2$ versus radius and $4πr2ψ2$ versus radius is to be drawn.

Concept introduction:

Quantum mechanics helps to understand the wave nature of particles. According to de Broglie, particles like electrons exhibit wave-like properties along with their particle properties.

The Schrodinger wave equation is represented as follows:

$H∧ψ=Eψ$

Here,

$E$ is the energy of the atom.

$H∧$ is the Hamiltonian operator.

$ψ$ is the wave function.

In the Schrodinger wave equation, $ψ$ represents the wave function. The wave function represents the wave associated with a particle. It has no physical significance.

However, the square of the wave function $(ψ2)$ gives the probability of finding an electron at a particular point and time. Thus it becomes possible to determine the region in an atom around the nucleus where there is a high possibility of finding an electron with a certain amount of energy. This region is called an atomic orbital.

The symbol $ψ2$ denotes the probability density. It gives the probability of finding an electron in an extremely small region in the atom

The expression to calculate the wave number of $1s$ the orbital is as follows:

$ψ=1π(1ao)32e−rao$ (1)

Here,

$ψ$ is the wave function.

$r$ is the distance from the nucleus.

Expert Solution & Answer

Answer to Problem 7.79P

The value of $ψ2$ , $ψ2$ and $4πr2ψ2$ for different value of $r$ are as follows:

$r(pm)ψψ24πr2ψ201.47×10−32.15×10−60500.570×10−30.325×10−61.02×10−21000.221×10−30.0491×10−60.0616×10−22000.0335×10−30.00112×10−60.0563×10−2$

The graph between $ψ$ versus radius, $ψ2$ versus radius and $4πr2ψ2$ versus r is as follows:

Connect 2-Year Access Card for Chemistry: The Molecular Nature of Matter and Change, Chapter 7, Problem 7.79P , additional homework tip 1

Explanation of Solution

Substitute $52.92 pm$ for $ao$ , $0$ for $r$ in the equation (1).

$ψ=1π(152.92 pm)32e−052.92 pm=1.465532×10−3=1.47×10−3$

The expression to calculate the probability density of electrons is as follows:

$Probability density of electron=ψ2$ (2)

Substitute $1.465532×10−3$ for $ψ$ in the equation (2).

$Probability density of electron=(1.465532×10−3)2=2.15×10−6 pm−3$

The expression to calculate the mass of an electron is as follows:

$Mass of electron=4πr2ψ2$ (3)

Substitute $2.15×10−6 pm−3$ for $ψ2$ and $0$ for $r$ in the equation (3).

$Mass of electron=4π(0)2(2.15×10−6 pm−3)=0$

Substitute $52.92 pm$ for $ao$ , $50 pm$ for $r$ in the equation (1).

$ψ=1π(152.92 pm)32e−50 pm52.92 pm=1.465532(e−50 pm52.92 pm)=5.69724×10−4$

Substitute $5.69724×10−4$ for $ψ$ in the equation (2).

$Probability density of electron=(5.69724×10−4)2=3.24585×10−7 pm−3$

Substitute $3.24585×10−7$ for $ψ2$ and $50 pm$ for $r$ in the equation (3).

$Mass of electron=4π(50 pm)2(3.24585×10−7)=1.0197×10−2 pm−1$

Substitute $52.92 pm$ for $ao$ , $100 pm$ for $r$ in the equation (1).

$ψ=1π(152.92 pm)32e−100 pm52.92 pm=1.465532(e−50 pm52.92 pm)=0.221×10−3$

Substitute $0.221×10−3$ for $ψ$ in the equation (2).

$Probability density of electron=(0.221×10−3)2=0.0491×10−6 pm−3$

Substitute $0.0491×10−6 pm−3$ for $ψ2$ and $50 pm$ for $r$ in the equation (3).

$Mass of electron=4π(50 pm)2(0.0491×10−6 pm−3)=0.616×10−2 pm−1$

Substitute $52.92 pm$ for $ao$ , $200 pm$ for $r$ in the equation (1).

$ψ=1π(152.92 pm)32e−200 pm52.92 pm=1.465532(e−50 pm52.92 pm)=0.0335×10−3$

Substitute $0.0335×10−3$ for $ψ$ in the equation (2).

$Probability density of electron=(0.0335×10−3)2=0.00112×10−6 pm−3$

Substitute $0.00112×10−6 pm−3$ for $ψ2$ and $200 pm$ for $r$ in the equation (3).

$Mass of electron=4π(50 pm)2(0.00112×10−6 pm−3)=0.0563×10−2 pm−1.$

The graph between $ψ$ versus radius, $ψ2$ versus radius and $4πr2ψ2$ versus r is as follows:

Connect 2-Year Access Card for Chemistry: The Molecular Nature of Matter and Change, Chapter 7, Problem 7.79P , additional homework tip 2

Conclusion

The graph between $ψ2$ versus radius plot has zero nodes.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 7, Problem 7.79P

Interpretation Introduction

Interpretation:

The value $ψ2$ , $ψ2$ and $4πr2ψ2$ for different value of $r$ are to be calculated. Also, the graph between $ψ$ versus radius, $ψ2$ versus radius and $4πr2ψ2$ versus radius is to be drawn.

Concept introduction:

Quantum mechanics helps to understand the wave nature of particles. According to de Broglie, particles like electrons exhibit wave-like properties along with their particle properties.

The Schrodinger wave equation is represented as follows:

$H∧ψ=Eψ$

Here,

$E$ is the energy of the atom.

$H∧$ is the Hamiltonian operator.

$ψ$ is the wave function.

In the Schrodinger wave equation, $ψ$ represents the wave function. The wave function represents the wave associated with a particle. It has no physical significance.

However, the square of the wave function $(ψ2)$ gives the probability of finding an electron at a particular point and time. Thus it becomes possible to determine the region in an atom around the nucleus where there is a high possibility of finding an electron with a certain amount of energy. This region is called an atomic orbital.

The symbol $ψ2$ denotes the probability density. It gives the probability of finding an electron in an extremely small region in the atom

The expression to calculate the wave number of $1s$ the orbital is as follows:

$ψ=1π(1ao)32e−rao$ (1)

Here,

$ψ$ is the wave function.

$r$ is the distance from the nucleus.

Expert Solution & Answer

Answer to Problem 7.79P

The value of $ψ2$ , $ψ2$ and $4πr2ψ2$ for different value of $r$ are as follows:

$r(pm)ψψ24πr2ψ201.47×10−32.15×10−60500.570×10−30.325×10−61.02×10−21000.221×10−30.0491×10−60.0616×10−22000.0335×10−30.00112×10−60.0563×10−2$

The graph between $ψ$ versus radius, $ψ2$ versus radius and $4πr2ψ2$ versus r is as follows:

Connect 2-Year Access Card for Chemistry: The Molecular Nature of Matter and Change, Chapter 7, Problem 7.79P , additional homework tip 1

Explanation of Solution

Substitute $52.92 pm$ for $ao$ , $0$ for $r$ in the equation (1).

$ψ=1π(152.92 pm)32e−052.92 pm=1.465532×10−3=1.47×10−3$

The expression to calculate the probability density of electrons is as follows:

$Probability density of electron=ψ2$ (2)

Substitute $1.465532×10−3$ for $ψ$ in the equation (2).

$Probability density of electron=(1.465532×10−3)2=2.15×10−6 pm−3$

The expression to calculate the mass of an electron is as follows:

$Mass of electron=4πr2ψ2$ (3)

Substitute $2.15×10−6 pm−3$ for $ψ2$ and $0$ for $r$ in the equation (3).

$Mass of electron=4π(0)2(2.15×10−6 pm−3)=0$

Substitute $52.92 pm$ for $ao$ , $50 pm$ for $r$ in the equation (1).

$ψ=1π(152.92 pm)32e−50 pm52.92 pm=1.465532(e−50 pm52.92 pm)=5.69724×10−4$

Substitute $5.69724×10−4$ for $ψ$ in the equation (2).

$Probability density of electron=(5.69724×10−4)2=3.24585×10−7 pm−3$

Substitute $3.24585×10−7$ for $ψ2$ and $50 pm$ for $r$ in the equation (3).

$Mass of electron=4π(50 pm)2(3.24585×10−7)=1.0197×10−2 pm−1$

Substitute $52.92 pm$ for $ao$ , $100 pm$ for $r$ in the equation (1).

$ψ=1π(152.92 pm)32e−100 pm52.92 pm=1.465532(e−50 pm52.92 pm)=0.221×10−3$

Substitute $0.221×10−3$ for $ψ$ in the equation (2).

$Probability density of electron=(0.221×10−3)2=0.0491×10−6 pm−3$

Substitute $0.0491×10−6 pm−3$ for $ψ2$ and $50 pm$ for $r$ in the equation (3).

$Mass of electron=4π(50 pm)2(0.0491×10−6 pm−3)=0.616×10−2 pm−1$

Substitute $52.92 pm$ for $ao$ , $200 pm$ for $r$ in the equation (1).

$ψ=1π(152.92 pm)32e−200 pm52.92 pm=1.465532(e−50 pm52.92 pm)=0.0335×10−3$

Substitute $0.0335×10−3$ for $ψ$ in the equation (2).

$Probability density of electron=(0.0335×10−3)2=0.00112×10−6 pm−3$

Substitute $0.00112×10−6 pm−3$ for $ψ2$ and $200 pm$ for $r$ in the equation (3).

$Mass of electron=4π(50 pm)2(0.00112×10−6 pm−3)=0.0563×10−2 pm−1.$

The graph between $ψ$ versus radius, $ψ2$ versus radius and $4πr2ψ2$ versus r is as follows:

Connect 2-Year Access Card for Chemistry: The Molecular Nature of Matter and Change, Chapter 7, Problem 7.79P , additional homework tip 2

Conclusion

The graph between $ψ2$ versus radius plot has zero nodes.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 7, Problem 7.79P

Interpretation Introduction

Interpretation:

The value $ψ2$ , $ψ2$ and $4πr2ψ2$ for different value of $r$ are to be calculated. Also, the graph between $ψ$ versus radius, $ψ2$ versus radius and $4πr2ψ2$ versus radius is to be drawn.

Concept introduction:

Quantum mechanics helps to understand the wave nature of particles. According to de Broglie, particles like electrons exhibit wave-like properties along with their particle properties.

The Schrodinger wave equation is represented as follows:

$H∧ψ=Eψ$

Here,

$E$ is the energy of the atom.

$H∧$ is the Hamiltonian operator.

$ψ$ is the wave function.

In the Schrodinger wave equation, $ψ$ represents the wave function. The wave function represents the wave associated with a particle. It has no physical significance.

However, the square of the wave function $(ψ2)$ gives the probability of finding an electron at a particular point and time. Thus it becomes possible to determine the region in an atom around the nucleus where there is a high possibility of finding an electron with a certain amount of energy. This region is called an atomic orbital.

The symbol $ψ2$ denotes the probability density. It gives the probability of finding an electron in an extremely small region in the atom

The expression to calculate the wave number of $1s$ the orbital is as follows:

$ψ=1π(1ao)32e−rao$ (1)

Here,

$ψ$ is the wave function.

$r$ is the distance from the nucleus.

Expert Solution & Answer

Answer to Problem 7.79P

The value of $ψ2$ , $ψ2$ and $4πr2ψ2$ for different value of $r$ are as follows:

$r(pm)ψψ24πr2ψ201.47×10−32.15×10−60500.570×10−30.325×10−61.02×10−21000.221×10−30.0491×10−60.0616×10−22000.0335×10−30.00112×10−60.0563×10−2$

The graph between $ψ$ versus radius, $ψ2$ versus radius and $4πr2ψ2$ versus r is as follows:

Connect 2-Year Access Card for Chemistry: The Molecular Nature of Matter and Change, Chapter 7, Problem 7.79P , additional homework tip 1

Explanation of Solution

Substitute $52.92 pm$ for $ao$ , $0$ for $r$ in the equation (1).

$ψ=1π(152.92 pm)32e−052.92 pm=1.465532×10−3=1.47×10−3$

The expression to calculate the probability density of electrons is as follows:

$Probability density of electron=ψ2$ (2)

Substitute $1.465532×10−3$ for $ψ$ in the equation (2).

$Probability density of electron=(1.465532×10−3)2=2.15×10−6 pm−3$

The expression to calculate the mass of an electron is as follows:

$Mass of electron=4πr2ψ2$ (3)

Substitute $2.15×10−6 pm−3$ for $ψ2$ and $0$ for $r$ in the equation (3).

$Mass of electron=4π(0)2(2.15×10−6 pm−3)=0$

Substitute $52.92 pm$ for $ao$ , $50 pm$ for $r$ in the equation (1).

$ψ=1π(152.92 pm)32e−50 pm52.92 pm=1.465532(e−50 pm52.92 pm)=5.69724×10−4$

Substitute $5.69724×10−4$ for $ψ$ in the equation (2).

$Probability density of electron=(5.69724×10−4)2=3.24585×10−7 pm−3$

Substitute $3.24585×10−7$ for $ψ2$ and $50 pm$ for $r$ in the equation (3).

$Mass of electron=4π(50 pm)2(3.24585×10−7)=1.0197×10−2 pm−1$

Substitute $52.92 pm$ for $ao$ , $100 pm$ for $r$ in the equation (1).

$ψ=1π(152.92 pm)32e−100 pm52.92 pm=1.465532(e−50 pm52.92 pm)=0.221×10−3$

Substitute $0.221×10−3$ for $ψ$ in the equation (2).

$Probability density of electron=(0.221×10−3)2=0.0491×10−6 pm−3$

Substitute $0.0491×10−6 pm−3$ for $ψ2$ and $50 pm$ for $r$ in the equation (3).

$Mass of electron=4π(50 pm)2(0.0491×10−6 pm−3)=0.616×10−2 pm−1$

Substitute $52.92 pm$ for $ao$ , $200 pm$ for $r$ in the equation (1).

$ψ=1π(152.92 pm)32e−200 pm52.92 pm=1.465532(e−50 pm52.92 pm)=0.0335×10−3$

Substitute $0.0335×10−3$ for $ψ$ in the equation (2).

$Probability density of electron=(0.0335×10−3)2=0.00112×10−6 pm−3$

Substitute $0.00112×10−6 pm−3$ for $ψ2$ and $200 pm$ for $r$ in the equation (3).

$Mass of electron=4π(50 pm)2(0.00112×10−6 pm−3)=0.0563×10−2 pm−1.$

The graph between $ψ$ versus radius, $ψ2$ versus radius and $4πr2ψ2$ versus r is as follows:

Connect 2-Year Access Card for Chemistry: The Molecular Nature of Matter and Change, Chapter 7, Problem 7.79P , additional homework tip 2

Conclusion

The graph between $ψ2$ versus radius plot has zero nodes.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 7, Problem 7.79P

Interpretation Introduction

Interpretation:

The value $ψ2$ , $ψ2$ and $4πr2ψ2$ for different value of $r$ are to be calculated. Also, the graph between $ψ$ versus radius, $ψ2$ versus radius and $4πr2ψ2$ versus radius is to be drawn.

Concept introduction:

Quantum mechanics helps to understand the wave nature of particles. According to de Broglie, particles like electrons exhibit wave-like properties along with their particle properties.

The Schrodinger wave equation is represented as follows:

$H∧ψ=Eψ$

Here,

$E$ is the energy of the atom.

$H∧$ is the Hamiltonian operator.

$ψ$ is the wave function.

In the Schrodinger wave equation, $ψ$ represents the wave function. The wave function represents the wave associated with a particle. It has no physical significance.

However, the square of the wave function $(ψ2)$ gives the probability of finding an electron at a particular point and time. Thus it becomes possible to determine the region in an atom around the nucleus where there is a high possibility of finding an electron with a certain amount of energy. This region is called an atomic orbital.

The symbol $ψ2$ denotes the probability density. It gives the probability of finding an electron in an extremely small region in the atom

The expression to calculate the wave number of $1s$ the orbital is as follows:

$ψ=1π(1ao)32e−rao$ (1)

Here,

$ψ$ is the wave function.

$r$ is the distance from the nucleus.

Expert Solution & Answer

Answer to Problem 7.79P

The value of $ψ2$ , $ψ2$ and $4πr2ψ2$ for different value of $r$ are as follows:

$r(pm)ψψ24πr2ψ201.47×10−32.15×10−60500.570×10−30.325×10−61.02×10−21000.221×10−30.0491×10−60.0616×10−22000.0335×10−30.00112×10−60.0563×10−2$

The graph between $ψ$ versus radius, $ψ2$ versus radius and $4πr2ψ2$ versus r is as follows:

Connect 2-Year Access Card for Chemistry: The Molecular Nature of Matter and Change, Chapter 7, Problem 7.79P , additional homework tip 1

Explanation of Solution

Substitute $52.92 pm$ for $ao$ , $0$ for $r$ in the equation (1).

$ψ=1π(152.92 pm)32e−052.92 pm=1.465532×10−3=1.47×10−3$

The expression to calculate the probability density of electrons is as follows:

$Probability density of electron=ψ2$ (2)

Substitute $1.465532×10−3$ for $ψ$ in the equation (2).

$Probability density of electron=(1.465532×10−3)2=2.15×10−6 pm−3$

The expression to calculate the mass of an electron is as follows:

$Mass of electron=4πr2ψ2$ (3)

Substitute $2.15×10−6 pm−3$ for $ψ2$ and $0$ for $r$ in the equation (3).

$Mass of electron=4π(0)2(2.15×10−6 pm−3)=0$

Substitute $52.92 pm$ for $ao$ , $50 pm$ for $r$ in the equation (1).

$ψ=1π(152.92 pm)32e−50 pm52.92 pm=1.465532(e−50 pm52.92 pm)=5.69724×10−4$

Substitute $5.69724×10−4$ for $ψ$ in the equation (2).

$Probability density of electron=(5.69724×10−4)2=3.24585×10−7 pm−3$

Substitute $3.24585×10−7$ for $ψ2$ and $50 pm$ for $r$ in the equation (3).

$Mass of electron=4π(50 pm)2(3.24585×10−7)=1.0197×10−2 pm−1$

Substitute $52.92 pm$ for $ao$ , $100 pm$ for $r$ in the equation (1).

$ψ=1π(152.92 pm)32e−100 pm52.92 pm=1.465532(e−50 pm52.92 pm)=0.221×10−3$

Substitute $0.221×10−3$ for $ψ$ in the equation (2).

$Probability density of electron=(0.221×10−3)2=0.0491×10−6 pm−3$

Substitute $0.0491×10−6 pm−3$ for $ψ2$ and $50 pm$ for $r$ in the equation (3).

$Mass of electron=4π(50 pm)2(0.0491×10−6 pm−3)=0.616×10−2 pm−1$

Substitute $52.92 pm$ for $ao$ , $200 pm$ for $r$ in the equation (1).

$ψ=1π(152.92 pm)32e−200 pm52.92 pm=1.465532(e−50 pm52.92 pm)=0.0335×10−3$

Substitute $0.0335×10−3$ for $ψ$ in the equation (2).

$Probability density of electron=(0.0335×10−3)2=0.00112×10−6 pm−3$

Substitute $0.00112×10−6 pm−3$ for $ψ2$ and $200 pm$ for $r$ in the equation (3).

$Mass of electron=4π(50 pm)2(0.00112×10−6 pm−3)=0.0563×10−2 pm−1.$

The graph between $ψ$ versus radius, $ψ2$ versus radius and $4πr2ψ2$ versus r is as follows:

Connect 2-Year Access Card for Chemistry: The Molecular Nature of Matter and Change, Chapter 7, Problem 7.79P , additional homework tip 2

Conclusion

The graph between $ψ2$ versus radius plot has zero nodes.

Answer 6

Chapter 7, Problem 7.79P

Interpretation Introduction

Interpretation:

The value $ψ2$ , $ψ2$ and $4πr2ψ2$ for different value of $r$ are to be calculated. Also, the graph between $ψ$ versus radius, $ψ2$ versus radius and $4πr2ψ2$ versus radius is to be drawn.

Concept introduction:

Quantum mechanics helps to understand the wave nature of particles. According to de Broglie, particles like electrons exhibit wave-like properties along with their particle properties.

The Schrodinger wave equation is represented as follows:

$H∧ψ=Eψ$

Here,

$E$ is the energy of the atom.

$H∧$ is the Hamiltonian operator.

$ψ$ is the wave function.

In the Schrodinger wave equation, $ψ$ represents the wave function. The wave function represents the wave associated with a particle. It has no physical significance.

However, the square of the wave function $(ψ2)$ gives the probability of finding an electron at a particular point and time. Thus it becomes possible to determine the region in an atom around the nucleus where there is a high possibility of finding an electron with a certain amount of energy. This region is called an atomic orbital.

The symbol $ψ2$ denotes the probability density. It gives the probability of finding an electron in an extremely small region in the atom

The expression to calculate the wave number of $1s$ the orbital is as follows:

$ψ=1π(1ao)32e−rao$ (1)

Here,

$ψ$ is the wave function.

$r$ is the distance from the nucleus.

Expert Solution & Answer

Answer to Problem 7.79P

The value of $ψ2$ , $ψ2$ and $4πr2ψ2$ for different value of $r$ are as follows:

$r(pm)ψψ24πr2ψ201.47×10−32.15×10−60500.570×10−30.325×10−61.02×10−21000.221×10−30.0491×10−60.0616×10−22000.0335×10−30.00112×10−60.0563×10−2$

The graph between $ψ$ versus radius, $ψ2$ versus radius and $4πr2ψ2$ versus r is as follows:

Connect 2-Year Access Card for Chemistry: The Molecular Nature of Matter and Change, Chapter 7, Problem 7.79P , additional homework tip 1

Explanation of Solution

Substitute $52.92 pm$ for $ao$ , $0$ for $r$ in the equation (1).

$ψ=1π(152.92 pm)32e−052.92 pm=1.465532×10−3=1.47×10−3$

The expression to calculate the probability density of electrons is as follows:

$Probability density of electron=ψ2$ (2)

Substitute $1.465532×10−3$ for $ψ$ in the equation (2).

$Probability density of electron=(1.465532×10−3)2=2.15×10−6 pm−3$

The expression to calculate the mass of an electron is as follows:

$Mass of electron=4πr2ψ2$ (3)

Substitute $2.15×10−6 pm−3$ for $ψ2$ and $0$ for $r$ in the equation (3).

$Mass of electron=4π(0)2(2.15×10−6 pm−3)=0$

Substitute $52.92 pm$ for $ao$ , $50 pm$ for $r$ in the equation (1).

$ψ=1π(152.92 pm)32e−50 pm52.92 pm=1.465532(e−50 pm52.92 pm)=5.69724×10−4$

Substitute $5.69724×10−4$ for $ψ$ in the equation (2).

$Probability density of electron=(5.69724×10−4)2=3.24585×10−7 pm−3$

Substitute $3.24585×10−7$ for $ψ2$ and $50 pm$ for $r$ in the equation (3).

$Mass of electron=4π(50 pm)2(3.24585×10−7)=1.0197×10−2 pm−1$

Substitute $52.92 pm$ for $ao$ , $100 pm$ for $r$ in the equation (1).

$ψ=1π(152.92 pm)32e−100 pm52.92 pm=1.465532(e−50 pm52.92 pm)=0.221×10−3$

Substitute $0.221×10−3$ for $ψ$ in the equation (2).

$Probability density of electron=(0.221×10−3)2=0.0491×10−6 pm−3$

Substitute $0.0491×10−6 pm−3$ for $ψ2$ and $50 pm$ for $r$ in the equation (3).

$Mass of electron=4π(50 pm)2(0.0491×10−6 pm−3)=0.616×10−2 pm−1$

Substitute $52.92 pm$ for $ao$ , $200 pm$ for $r$ in the equation (1).

$ψ=1π(152.92 pm)32e−200 pm52.92 pm=1.465532(e−50 pm52.92 pm)=0.0335×10−3$

Substitute $0.0335×10−3$ for $ψ$ in the equation (2).

$Probability density of electron=(0.0335×10−3)2=0.00112×10−6 pm−3$

Substitute $0.00112×10−6 pm−3$ for $ψ2$ and $200 pm$ for $r$ in the equation (3).

$Mass of electron=4π(50 pm)2(0.00112×10−6 pm−3)=0.0563×10−2 pm−1.$

The graph between $ψ$ versus radius, $ψ2$ versus radius and $4πr2ψ2$ versus r is as follows:

Connect 2-Year Access Card for Chemistry: The Molecular Nature of Matter and Change, Chapter 7, Problem 7.79P , additional homework tip 2

Conclusion

The graph between $ψ2$ versus radius plot has zero nodes.

Answer 7

Chapter 7, Problem 7.79P

Interpretation Introduction

Interpretation:

The value $ψ2$ , $ψ2$ and $4πr2ψ2$ for different value of $r$ are to be calculated. Also, the graph between $ψ$ versus radius, $ψ2$ versus radius and $4πr2ψ2$ versus radius is to be drawn.

Concept introduction:

Quantum mechanics helps to understand the wave nature of particles. According to de Broglie, particles like electrons exhibit wave-like properties along with their particle properties.

The Schrodinger wave equation is represented as follows:

$H∧ψ=Eψ$

Here,

$E$ is the energy of the atom.

$H∧$ is the Hamiltonian operator.

$ψ$ is the wave function.

In the Schrodinger wave equation, $ψ$ represents the wave function. The wave function represents the wave associated with a particle. It has no physical significance.

However, the square of the wave function $(ψ2)$ gives the probability of finding an electron at a particular point and time. Thus it becomes possible to determine the region in an atom around the nucleus where there is a high possibility of finding an electron with a certain amount of energy. This region is called an atomic orbital.

The symbol $ψ2$ denotes the probability density. It gives the probability of finding an electron in an extremely small region in the atom

The expression to calculate the wave number of $1s$ the orbital is as follows:

$ψ=1π(1ao)32e−rao$ (1)

Here,

$ψ$ is the wave function.

$r$ is the distance from the nucleus.

Answer 8

Expert Solution & Answer

Answer to Problem 7.79P

The value of $ψ2$ , $ψ2$ and $4πr2ψ2$ for different value of $r$ are as follows:

$r(pm)ψψ24πr2ψ201.47×10−32.15×10−60500.570×10−30.325×10−61.02×10−21000.221×10−30.0491×10−60.0616×10−22000.0335×10−30.00112×10−60.0563×10−2$

The graph between $ψ$ versus radius, $ψ2$ versus radius and $4πr2ψ2$ versus r is as follows:

Connect 2-Year Access Card for Chemistry: The Molecular Nature of Matter and Change, Chapter 7, Problem 7.79P , additional homework tip 1

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Substitute $52.92 pm$ for $ao$ , $0$ for $r$ in the equation (1).

$ψ=1π(152.92 pm)32e−052.92 pm=1.465532×10−3=1.47×10−3$

The expression to calculate the probability density of electrons is as follows:

$Probability density of electron=ψ2$ (2)

Substitute $1.465532×10−3$ for $ψ$ in the equation (2).

$Probability density of electron=(1.465532×10−3)2=2.15×10−6 pm−3$

The expression to calculate the mass of an electron is as follows:

$Mass of electron=4πr2ψ2$ (3)

Substitute $2.15×10−6 pm−3$ for $ψ2$ and $0$ for $r$ in the equation (3).

$Mass of electron=4π(0)2(2.15×10−6 pm−3)=0$

Substitute $52.92 pm$ for $ao$ , $50 pm$ for $r$ in the equation (1).

$ψ=1π(152.92 pm)32e−50 pm52.92 pm=1.465532(e−50 pm52.92 pm)=5.69724×10−4$

Substitute $5.69724×10−4$ for $ψ$ in the equation (2).

$Probability density of electron=(5.69724×10−4)2=3.24585×10−7 pm−3$

Substitute $3.24585×10−7$ for $ψ2$ and $50 pm$ for $r$ in the equation (3).

$Mass of electron=4π(50 pm)2(3.24585×10−7)=1.0197×10−2 pm−1$

Substitute $52.92 pm$ for $ao$ , $100 pm$ for $r$ in the equation (1).

$ψ=1π(152.92 pm)32e−100 pm52.92 pm=1.465532(e−50 pm52.92 pm)=0.221×10−3$

Substitute $0.221×10−3$ for $ψ$ in the equation (2).

$Probability density of electron=(0.221×10−3)2=0.0491×10−6 pm−3$

Substitute $0.0491×10−6 pm−3$ for $ψ2$ and $50 pm$ for $r$ in the equation (3).

$Mass of electron=4π(50 pm)2(0.0491×10−6 pm−3)=0.616×10−2 pm−1$

Substitute $52.92 pm$ for $ao$ , $200 pm$ for $r$ in the equation (1).

$ψ=1π(152.92 pm)32e−200 pm52.92 pm=1.465532(e−50 pm52.92 pm)=0.0335×10−3$

Substitute $0.0335×10−3$ for $ψ$ in the equation (2).

$Probability density of electron=(0.0335×10−3)2=0.00112×10−6 pm−3$

Substitute $0.00112×10−6 pm−3$ for $ψ2$ and $200 pm$ for $r$ in the equation (3).

$Mass of electron=4π(50 pm)2(0.00112×10−6 pm−3)=0.0563×10−2 pm−1.$

The graph between $ψ$ versus radius, $ψ2$ versus radius and $4πr2ψ2$ versus r is as follows:

Connect 2-Year Access Card for Chemistry: The Molecular Nature of Matter and Change, Chapter 7, Problem 7.79P , additional homework tip 2

Answer 12

Conclusion

The graph between $ψ2$ versus radius plot has zero nodes.

Answer 13

Ch. 7.1 - Some diamonds appear yellow because they contain...Ch. 7.1 - Prob. 7.1BFP Ch. 7.1 - Prob. 7.2AFP Ch. 7.1 - Prob. 7.2BFP Ch. 7.2 - Prob. 7.3AFP Ch. 7.2 - Prob. 7.3BFP Ch. 7.2 - Prob. B7.1P Ch. 7.2 - Prob. B7.2P Ch. 7.3 - Prob. 7.4AFP Ch. 7.3 - Prob. 7.4BFP

Concept explainers

Videos

Answer to Problem 7.79P

Explanation of Solution

Want to see more full solutions like this?

Chapter 7 Solutions