Whether the wave function for 2 p state is normalized or not.

Question

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Answer 1

Question

Chapter 28, Problem 41E

To determine

Whether the wave function for $2p$ state is normalized or not.

Expert Solution & Answer

Answer to Problem 41E

The wave function for $2p$ state is normalized.

Explanation of Solution

Write the expression of the wave function for $2p$ state of hydrogen atom.

$Ψ(r) = 1π1/2(2a0)3/2ra0e−r/2a0Cosθ$ (I)

Here, $a0$ is first Bohr’s radius, $r$ is the atomic radius and $θ$ is the azimuthal angle.

Write the expression for the condition of normalization of the wave function.

$∫|Ψ(r)|2d3r = 1$ (II)

Conclusion:

Substitute $1π1/2(2a0)3/2ra0e−r/2a0Cosθ$ for $Ψ(r)$ on the left-hand side of expression (II).

$∫|Ψ(r)|2d3r = ∫0∞|1π1/2(2a0)3/2ra0e−r/2a0Cosθ|2r2dr∫0πSinθdθ∫02πdϕ$

Simplify the above expression.

$∫|Ψ(r)|2d3r = 132πa05∫0∞r4e−r/a0dr∫0πCos2θSinθdθ ∫02πdϕ$

Evaluate the last integral on the right-hand side of above expression.

$∫02πdϕ = 2π$

Evaluate the second integral on the right-hand side of expression.

$∫0πCos2θSinθdθ = − ∫0πCos2θd(Cosθ)= − 13[Cos3θ]0π= −13[−1 −1]= 23$

Substitute $x$ for $r/a0$ in the first integral on the right-hand side of expression.

$∫0∞r4e−r/a0dr =∫0∞(a0x)4e−x(a0dx)= (a05)∫0∞x4e−xdx$

Use the identity given below to evaluate the integral on the right-hand side of above expression.

$∫0∞xne−xdx = Γ(n)$

Simplify the expression.

$∫0∞r4e−r/a0dr = (a05)Γ(4)= 24a05$

Substitute $24a05$ for $∫0∞r4e−r/a0dr$ , $23$ for $∫0πCos2θSinθdθ$ and $2π$ for $∫02πdϕ$ in the wave function expression.

$∫|Ψ(r)|2d3r = 132πa05(24a05)(23)(2π)= 1$

Thus, the wave function for $2p$ state is normalized.

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Answer 2

Question

Chapter 28, Problem 41E

To determine

Whether the wave function for $2p$ state is normalized or not.

Expert Solution & Answer

Answer to Problem 41E

The wave function for $2p$ state is normalized.

Explanation of Solution

Write the expression of the wave function for $2p$ state of hydrogen atom.

$Ψ(r) = 1π1/2(2a0)3/2ra0e−r/2a0Cosθ$ (I)

Here, $a0$ is first Bohr’s radius, $r$ is the atomic radius and $θ$ is the azimuthal angle.

Write the expression for the condition of normalization of the wave function.

$∫|Ψ(r)|2d3r = 1$ (II)

Conclusion:

Substitute $1π1/2(2a0)3/2ra0e−r/2a0Cosθ$ for $Ψ(r)$ on the left-hand side of expression (II).

$∫|Ψ(r)|2d3r = ∫0∞|1π1/2(2a0)3/2ra0e−r/2a0Cosθ|2r2dr∫0πSinθdθ∫02πdϕ$

Simplify the above expression.

$∫|Ψ(r)|2d3r = 132πa05∫0∞r4e−r/a0dr∫0πCos2θSinθdθ ∫02πdϕ$

Evaluate the last integral on the right-hand side of above expression.

$∫02πdϕ = 2π$

Evaluate the second integral on the right-hand side of expression.

$∫0πCos2θSinθdθ = − ∫0πCos2θd(Cosθ)= − 13[Cos3θ]0π= −13[−1 −1]= 23$

Substitute $x$ for $r/a0$ in the first integral on the right-hand side of expression.

$∫0∞r4e−r/a0dr =∫0∞(a0x)4e−x(a0dx)= (a05)∫0∞x4e−xdx$

Use the identity given below to evaluate the integral on the right-hand side of above expression.

$∫0∞xne−xdx = Γ(n)$

Simplify the expression.

$∫0∞r4e−r/a0dr = (a05)Γ(4)= 24a05$

Substitute $24a05$ for $∫0∞r4e−r/a0dr$ , $23$ for $∫0πCos2θSinθdθ$ and $2π$ for $∫02πdϕ$ in the wave function expression.

$∫|Ψ(r)|2d3r = 132πa05(24a05)(23)(2π)= 1$

Thus, the wave function for $2p$ state is normalized.

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Answer 3

Question

Chapter 28, Problem 41E

To determine

Whether the wave function for $2p$ state is normalized or not.

Expert Solution & Answer

Answer to Problem 41E

The wave function for $2p$ state is normalized.

Explanation of Solution

Write the expression of the wave function for $2p$ state of hydrogen atom.

$Ψ(r) = 1π1/2(2a0)3/2ra0e−r/2a0Cosθ$ (I)

Here, $a0$ is first Bohr’s radius, $r$ is the atomic radius and $θ$ is the azimuthal angle.

Write the expression for the condition of normalization of the wave function.

$∫|Ψ(r)|2d3r = 1$ (II)

Conclusion:

Substitute $1π1/2(2a0)3/2ra0e−r/2a0Cosθ$ for $Ψ(r)$ on the left-hand side of expression (II).

$∫|Ψ(r)|2d3r = ∫0∞|1π1/2(2a0)3/2ra0e−r/2a0Cosθ|2r2dr∫0πSinθdθ∫02πdϕ$

Simplify the above expression.

$∫|Ψ(r)|2d3r = 132πa05∫0∞r4e−r/a0dr∫0πCos2θSinθdθ ∫02πdϕ$

Evaluate the last integral on the right-hand side of above expression.

$∫02πdϕ = 2π$

Evaluate the second integral on the right-hand side of expression.

$∫0πCos2θSinθdθ = − ∫0πCos2θd(Cosθ)= − 13[Cos3θ]0π= −13[−1 −1]= 23$

Substitute $x$ for $r/a0$ in the first integral on the right-hand side of expression.

$∫0∞r4e−r/a0dr =∫0∞(a0x)4e−x(a0dx)= (a05)∫0∞x4e−xdx$

Use the identity given below to evaluate the integral on the right-hand side of above expression.

$∫0∞xne−xdx = Γ(n)$

Simplify the expression.

$∫0∞r4e−r/a0dr = (a05)Γ(4)= 24a05$

Substitute $24a05$ for $∫0∞r4e−r/a0dr$ , $23$ for $∫0πCos2θSinθdθ$ and $2π$ for $∫02πdϕ$ in the wave function expression.

$∫|Ψ(r)|2d3r = 132πa05(24a05)(23)(2π)= 1$

Thus, the wave function for $2p$ state is normalized.

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Answer 4

Question

Chapter 28, Problem 41E

To determine

Whether the wave function for $2p$ state is normalized or not.

Expert Solution & Answer

Answer to Problem 41E

The wave function for $2p$ state is normalized.

Explanation of Solution

Write the expression of the wave function for $2p$ state of hydrogen atom.

$Ψ(r) = 1π1/2(2a0)3/2ra0e−r/2a0Cosθ$ (I)

Here, $a0$ is first Bohr’s radius, $r$ is the atomic radius and $θ$ is the azimuthal angle.

Write the expression for the condition of normalization of the wave function.

$∫|Ψ(r)|2d3r = 1$ (II)

Conclusion:

Substitute $1π1/2(2a0)3/2ra0e−r/2a0Cosθ$ for $Ψ(r)$ on the left-hand side of expression (II).

$∫|Ψ(r)|2d3r = ∫0∞|1π1/2(2a0)3/2ra0e−r/2a0Cosθ|2r2dr∫0πSinθdθ∫02πdϕ$

Simplify the above expression.

$∫|Ψ(r)|2d3r = 132πa05∫0∞r4e−r/a0dr∫0πCos2θSinθdθ ∫02πdϕ$

Evaluate the last integral on the right-hand side of above expression.

$∫02πdϕ = 2π$

Evaluate the second integral on the right-hand side of expression.

$∫0πCos2θSinθdθ = − ∫0πCos2θd(Cosθ)= − 13[Cos3θ]0π= −13[−1 −1]= 23$

Substitute $x$ for $r/a0$ in the first integral on the right-hand side of expression.

$∫0∞r4e−r/a0dr =∫0∞(a0x)4e−x(a0dx)= (a05)∫0∞x4e−xdx$

Use the identity given below to evaluate the integral on the right-hand side of above expression.

$∫0∞xne−xdx = Γ(n)$

Simplify the expression.

$∫0∞r4e−r/a0dr = (a05)Γ(4)= 24a05$

Substitute $24a05$ for $∫0∞r4e−r/a0dr$ , $23$ for $∫0πCos2θSinθdθ$ and $2π$ for $∫02πdϕ$ in the wave function expression.

$∫|Ψ(r)|2d3r = 132πa05(24a05)(23)(2π)= 1$

Thus, the wave function for $2p$ state is normalized.

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Answer 5

Chapter 28, Problem 41E

To determine

Whether the wave function for $2p$ state is normalized or not.

Expert Solution & Answer

Answer to Problem 41E

The wave function for $2p$ state is normalized.

Explanation of Solution

Write the expression of the wave function for $2p$ state of hydrogen atom.

$Ψ(r) = 1π1/2(2a0)3/2ra0e−r/2a0Cosθ$ (I)

Here, $a0$ is first Bohr’s radius, $r$ is the atomic radius and $θ$ is the azimuthal angle.

Write the expression for the condition of normalization of the wave function.

$∫|Ψ(r)|2d3r = 1$ (II)

Conclusion:

Substitute $1π1/2(2a0)3/2ra0e−r/2a0Cosθ$ for $Ψ(r)$ on the left-hand side of expression (II).

$∫|Ψ(r)|2d3r = ∫0∞|1π1/2(2a0)3/2ra0e−r/2a0Cosθ|2r2dr∫0πSinθdθ∫02πdϕ$

Simplify the above expression.

$∫|Ψ(r)|2d3r = 132πa05∫0∞r4e−r/a0dr∫0πCos2θSinθdθ ∫02πdϕ$

Evaluate the last integral on the right-hand side of above expression.

$∫02πdϕ = 2π$

Evaluate the second integral on the right-hand side of expression.

$∫0πCos2θSinθdθ = − ∫0πCos2θd(Cosθ)= − 13[Cos3θ]0π= −13[−1 −1]= 23$

Substitute $x$ for $r/a0$ in the first integral on the right-hand side of expression.

$∫0∞r4e−r/a0dr =∫0∞(a0x)4e−x(a0dx)= (a05)∫0∞x4e−xdx$

Use the identity given below to evaluate the integral on the right-hand side of above expression.

$∫0∞xne−xdx = Γ(n)$

Simplify the expression.

$∫0∞r4e−r/a0dr = (a05)Γ(4)= 24a05$

Substitute $24a05$ for $∫0∞r4e−r/a0dr$ , $23$ for $∫0πCos2θSinθdθ$ and $2π$ for $∫02πdϕ$ in the wave function expression.

$∫|Ψ(r)|2d3r = 132πa05(24a05)(23)(2π)= 1$

Thus, the wave function for $2p$ state is normalized.

Answer 6

Chapter 28, Problem 41E

To determine

Whether the wave function for $2p$ state is normalized or not.

Expert Solution & Answer

Answer to Problem 41E

The wave function for $2p$ state is normalized.

Explanation of Solution

Write the expression of the wave function for $2p$ state of hydrogen atom.

$Ψ(r) = 1π1/2(2a0)3/2ra0e−r/2a0Cosθ$ (I)

Here, $a0$ is first Bohr’s radius, $r$ is the atomic radius and $θ$ is the azimuthal angle.

Write the expression for the condition of normalization of the wave function.

$∫|Ψ(r)|2d3r = 1$ (II)

Conclusion:

Substitute $1π1/2(2a0)3/2ra0e−r/2a0Cosθ$ for $Ψ(r)$ on the left-hand side of expression (II).

$∫|Ψ(r)|2d3r = ∫0∞|1π1/2(2a0)3/2ra0e−r/2a0Cosθ|2r2dr∫0πSinθdθ∫02πdϕ$

Simplify the above expression.

$∫|Ψ(r)|2d3r = 132πa05∫0∞r4e−r/a0dr∫0πCos2θSinθdθ ∫02πdϕ$

Evaluate the last integral on the right-hand side of above expression.

$∫02πdϕ = 2π$

Evaluate the second integral on the right-hand side of expression.

$∫0πCos2θSinθdθ = − ∫0πCos2θd(Cosθ)= − 13[Cos3θ]0π= −13[−1 −1]= 23$

Substitute $x$ for $r/a0$ in the first integral on the right-hand side of expression.

$∫0∞r4e−r/a0dr =∫0∞(a0x)4e−x(a0dx)= (a05)∫0∞x4e−xdx$

Use the identity given below to evaluate the integral on the right-hand side of above expression.

$∫0∞xne−xdx = Γ(n)$

Simplify the expression.

$∫0∞r4e−r/a0dr = (a05)Γ(4)= 24a05$

Substitute $24a05$ for $∫0∞r4e−r/a0dr$ , $23$ for $∫0πCos2θSinθdθ$ and $2π$ for $∫02πdϕ$ in the wave function expression.

$∫|Ψ(r)|2d3r = 132πa05(24a05)(23)(2π)= 1$

Thus, the wave function for $2p$ state is normalized.

Answer 7

Chapter 28, Problem 41E

To determine

Whether the wave function for $2p$ state is normalized or not.

Answer 8

Expert Solution & Answer

Answer to Problem 41E

The wave function for $2p$ state is normalized.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Write the expression of the wave function for $2p$ state of hydrogen atom.

$Ψ(r) = 1π1/2(2a0)3/2ra0e−r/2a0Cosθ$ (I)

Here, $a0$ is first Bohr’s radius, $r$ is the atomic radius and $θ$ is the azimuthal angle.

Write the expression for the condition of normalization of the wave function.

$∫|Ψ(r)|2d3r = 1$ (II)

Conclusion:

Substitute $1π1/2(2a0)3/2ra0e−r/2a0Cosθ$ for $Ψ(r)$ on the left-hand side of expression (II).

$∫|Ψ(r)|2d3r = ∫0∞|1π1/2(2a0)3/2ra0e−r/2a0Cosθ|2r2dr∫0πSinθdθ∫02πdϕ$

Simplify the above expression.

$∫|Ψ(r)|2d3r = 132πa05∫0∞r4e−r/a0dr∫0πCos2θSinθdθ ∫02πdϕ$

Evaluate the last integral on the right-hand side of above expression.

$∫02πdϕ = 2π$

Evaluate the second integral on the right-hand side of expression.

$∫0πCos2θSinθdθ = − ∫0πCos2θd(Cosθ)= − 13[Cos3θ]0π= −13[−1 −1]= 23$

Substitute $x$ for $r/a0$ in the first integral on the right-hand side of expression.

$∫0∞r4e−r/a0dr =∫0∞(a0x)4e−x(a0dx)= (a05)∫0∞x4e−xdx$

Use the identity given below to evaluate the integral on the right-hand side of above expression.

$∫0∞xne−xdx = Γ(n)$

Simplify the expression.

$∫0∞r4e−r/a0dr = (a05)Γ(4)= 24a05$

Substitute $24a05$ for $∫0∞r4e−r/a0dr$ , $23$ for $∫0πCos2θSinθdθ$ and $2π$ for $∫02πdϕ$ in the wave function expression.

$∫|Ψ(r)|2d3r = 132πa05(24a05)(23)(2π)= 1$

Thus, the wave function for $2p$ state is normalized.

Answer 12

Ch. 28 - Prob. 1RQ Ch. 28 - Prob. 2RQ Ch. 28 - Prob. 3RQ Ch. 28 - Prob. 4RQ Ch. 28 - Prob. 5RQ Ch. 28 - Prob. 6RQ Ch. 28 - Prob. 7RQ Ch. 28 - Prob. 8RQ Ch. 28 - Prob. 9RQ Ch. 28 - Prob. 10RQ

Whether the wave function for 2 p state is normalized or not.

Whether the wave function for 2p<m:math><m:mrow><m:mn>2</m:mn><m:mi>p</m:mi></m:mrow></m:math> state is normalized or not.

Answer to Problem 41E

Explanation of Solution

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Chapter 28 Solutions

Whether the wave function for $2p$ state is normalized or not.