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 10, Problem 10.26E

For a particle in a state having the wavefunction

Ψ = 2 a sin π x a in the range x = 0 to a , what is the probability that the particle exists in the following intervals?

(a) x = 0 to 0.02 a (b) x = 0.24 a to 0.26 a

(c) x = 0.49 a to 0.51 a (d) x = 0.74 a to 0.76 a

(e) x = 0.98 a to 1.00 a

Plot the probabilities versus x . What does your plot illustrate about the probability?

Expert Solution
Check Mark
Interpretation Introduction

(a)

Interpretation:

The probability for the particle having wavefunction Ψ=2asinπxa in the range of x=0to0.02a is to be calculated.

Concept introduction:

For the normalization of the wavefunction, the wavefunction is integrated as a product of its conjugate over the entire limits. It is expressed by the equation as given below.

0(NΨ)(NΨ*)=1

Where,

N is the normalization constant

Ψ* is the conjugate of the wavefunction

Ψ is the wavefunction

Answer to Problem 10.26E

The probability for the particle having wavefunction Ψ=2asinπxa in the range of x=0to0.02a is 5.57×105.

Explanation of Solution

For the probability of the wavefunction the expression is as follows.

0(NΨ)(NΨ*)=P

Where,

N is the normalization constant

Ψ* is the conjugate of the wavefunction

Ψ is the wavefunction

P is the probability

Substitute the values in the above equation as follows.

P=00.02a(Ψ)(Ψ*)dx=00.02a(2asinπxa)(2asinπxa)dx=2a00.02a(sinπxa)(sinπxa)dx=2a00.02a(sinπxa)2dx

The above expression is simplified as follows.

=2a00.02a(sin2πxa)dx=2a×[x2a4πsin2πxa]00.02a=2a×[(0.02a2a4πsin2π(0.02a)a)0]=2a×(0.02a2a4πsin0.04π)

=2a×(0.02a2a12.56sin(0.1256))=2a×(0.02a20.12527a12.56)=2a×0.0007a25.12=5.57×105

Conclusion

The probability for the particle having wavefunction Ψ=2asinπxa in the range of x=0to0.02a is calculated as 5.57×105.

Expert Solution
Check Mark
Interpretation Introduction

(b)

Interpretation:

The probability for the particle having wavefunction Ψ=2asinπxa in the range of x=0.24ato0.26a is to be calculated.

Concept introduction:

For the normalization of the wavefunction, the wavefunction is integrated as a product of its conjugate over the entire limits. It is expressed by the equation as given below.

0(NΨ)(NΨ*)=1

Where,

N is the normalization constant

Ψ* is the conjugate of the wavefunction

Ψ is the wavefunction

Answer to Problem 10.26E

The probability for the particle having wavefunction Ψ=2asinπxa in the range of x=0.24ato0.26a is 0.0199.

Explanation of Solution

For the probability of the wavefunction the expression is as follows.

0(NΨ)(NΨ*)=P

Where,

N is the normalization constant

Ψ* is the conjugate of the wavefunction

Ψ is the wavefunction

P is the probability

Substitute the values in the above equation as follows.

P=0.24a0.26a(Ψ)(Ψ*)dx=0.24a0.26a(2asinπxa)(2asinπxa)dx=2a0.24a0.26a(sinπxa)(sinπxa)dx=2a0.24a0.26a(sinπxa)2dx

The above expression is simplified as follows.

=2a0.24a0.26a(sin2πxa)dx=2a×[x2a4πsin2πxa]0.24a0.26a=2a×[(0.26a2a4πsin2π(0.26a)a)(0.24a2a4πsin2π(0.24a)a)]=2a×(0.26a20.24a2a4πsin0.52π+a4πsin0.48π)

==2a×(0.26a20.24a2a4πsin0.52π+a4πsin0.48π)=2a×(0.02a20.0001a12.56)=2a×0.251a25.12=0.0199

Conclusion

The probability for the particle having wavefunction Ψ=2asinπxa in the range of x=0.24ato0.26a is calculated as 0.0199.

Expert Solution
Check Mark
Interpretation Introduction

(c)

Interpretation:

The probability for the particle having wavefunction Ψ=2asinπxa in the range of x=0.49ato0.51a is to be calculated.

Concept introduction:

For the normalization of the wavefunction, the wavefunction is integrated as a product of its conjugate over the entire limits. It is expressed by the equation as given below.

0(NΨ)(NΨ*)=1

Where,

N is the normalization constant

Ψ* is the conjugate of the wavefunction

Ψ is the wavefunction

Answer to Problem 10.26E

The probability for the particle having wavefunction Ψ=2asinπxa in the range of x=0.49ato0.51a is 0.0399.

Explanation of Solution

For the probability of the wavefunction the expression is as follows.

0(NΨ)(NΨ*)=P

Where,

N is the normalization constant

Ψ* is the conjugate of the wavefunction

Ψ is the wavefunction

P is the probability

Substitute the values in the above equation as follows.

P=0.49a0.51a(Ψ)(Ψ*)dx=0.49a0.51a(2asinπxa)(2asinπxa)dx=2a0.49a0.51a(sinπxa)(sinπxa)dx=2a0.49a0.51a(sinπxa)2dx

The above expression is simplified as follows.

=2a0.49a0.51a(sin2πxa)dx=2a×[x2a4πsin2πxa]0.49a0.51a=2a×[(0.51a2a4πsin2π(0.51a)a)(0.49a2a4πsin2π(0.49a)a)]=2a×(0.51a20.49a2a4πsin1.02π+a4πsin0.98π)

=2a×((0.51a20.49a2)a4πsin1.02π+a4πsin0.98π)=2a×(0.02a2+0.125517a12.56)=2a×0.502234a25.12=0.0399

Conclusion

The probability for the particle having wavefunction Ψ=2asinπxa in the range of x=0.49ato0.51a is calculated as 0.0399.

Expert Solution
Check Mark
Interpretation Introduction

(d)

Interpretation:

The probability for the particle having wavefunction Ψ=2asinπxa in the range of x=0.74ato0.76a is to be calculated.

Concept introduction:

For the normalization of the wavefunction, the wavefunction is integrated as a product of its conjugate over the entire limits. It is expressed by the equation as given below.

0(NΨ)(NΨ*)=1

Where,

N is the normalization constant

Ψ* is the conjugate of the wavefunction

Ψ is the wavefunction

Answer to Problem 10.26E

The probability for the particle having wavefunction Ψ=2asinπxa in the range of x=0.74ato0.76a is 0.0204.

Explanation of Solution

For the probability of the wavefunction the expression is as follows.

0(NΨ)(NΨ*)=P

Where,

N is the normalization constant

Ψ* is the conjugate of the wavefunction

Ψ is the wavefunction

P is the probability

Substitute the values in the above equation as follows.

P=0.74a0.76a(Ψ)(Ψ*)dx=0.74a0.76a(2asinπxa)(2asinπxa)dx=2a0.74a0.76a(sinπxa)(sinπxa)dx=2a0.74a0.76a(sinπxa)2dx

The above expression is simplified as follows.

=2a0.74a0.76a(sin2πxa)dx=2a×[x2a4πsin2πxa]0.74a0.76a=2a×[(0.76a2a4πsin2π(0.76a)a)(0.74a2a4πsin2π(0.74a)a)]=2a×(0.76a20.74a2a4πsin1.52π+a4πsin1.48π)

=2a×((0.76a20.74a2)a4πsin1.52π+a4πsin1.48π)=2a×(0.02a2+0.003a12.56)=2a×0.2572a25.12=0.0204

Conclusion

The probability for the particle having wavefunction Ψ=2asinπxa in the range of x=0.74ato0.76a is calculated as 0.0204.

Expert Solution
Check Mark
Interpretation Introduction

(e)

Interpretation:

The probability for the particle having wavefunction Ψ=2asinπxa in the range of x=0.98ato1.00a is to be calculated. The graph of probabilities versus x is to be plotted.

Concept introduction:

For the normalization of the wavefunction, the wavefunction is integrated as a product of its conjugate over the entire limits. It is expressed by the equation as given below.

0(NΨ)(NΨ*)=1

Where,

N is the normalization constant

Ψ* is the conjugate of the wavefunction

Ψ is the wavefunction

Answer to Problem 10.26E

The probability for the particle having wavefunction Ψ=2asinπxa in the range of x=0.98ato1.00a is 5×105.The graph of probabilities versus x is given below.

Physical Chemistry, Chapter 10, Problem 10.26E , additional homework tip  1

Explanation of Solution

For the probability of the wavefunction the expression is as follows.

0(NΨ)(NΨ*)=P

Where,

N is the normalization constant

Ψ* is the conjugate of the wavefunction

Ψ is the wavefunction

P is the probability

Substitute the values in the above equation as follows.

P=0.98a1a(Ψ)(Ψ*)dx=0.98a1a(2asinπxa)(2asinπxa)dx=2a0.98a1a(sinπxa)(sinπxa)dx=2a0.98a1a(sinπxa)2dx

The above expression is simplified as follows.

=2a0.98a1a(sin2πxa)dx=2a×[x2a4πsin2πxa]0.98a1a=2a×[(a2a4πsin2π(a)a)(0.98a2a4πsin2π(0.98a)a)]=2a×(a20.98a2a4πsin2π+a4πsin1.96π)

=2a×((a20.98a2)a4πsin2π+a4πsin1.96π)=2a×(0.02a20.125244a12.56)=2a×(0.000712a25.12)=5×105

Theplot the probabilities versus x is given in figure 1.

Physical Chemistry, Chapter 10, Problem 10.26E , additional homework tip  2

Figure 1

The plot shows the probability for the given wave function. According to this plot, the probability of finding the particle is maximum in the range of x=0.49ato0.51a. Also, the plot gives the curve for the sine function as given.

Conclusion

The probability for the particle having wavefunction Ψ=2asinπxa in the range of x=0.98ato1.00a is calculated as 5×105.

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