Question

1C.5 (a) Use the Living Graphs on the Web site for this book

to plot the particle-in-a-box wavefunction for n = 2 and L = 1 m. (b) How many nodes does the wavefunction have? Where do these nodes occur? (c) Repeat parts (a) and (b) for n = 3. (d) What general conclusion can you draw about the relation between n and the number of nodes present in a wavefunction? (e) Convert the n = 2 plot to a probability density distribution: at what values of x is it most likely to find the particle? (f) Repeat part (e) for n = 3.

1C.6 Verify the conclusion in part (d) of Exercise 1C.5 by

plotting the wavefunction for n = 4 and determining the number of nodes.

Answer 1

ANSWER - QUESTION 1C.5(a):

The equation that define the allowed wavefunctions for a particle-in-a-box is:

For n=2 and L = 1.0 m, the equation changes to

x (m)	Ψ (x) (m-1/2)
0
1/4 = 0.25
2/4 = 0.5
3/4 = 0.75
1

Then the plot of Ψ (x) vs. x gives:

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ANSWER - QUESTION 1C.5(b):

According to the graph, at n=2 and L = 1 m, the wavefunction equation has only 1 node at x = 1/2

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ANSWER - QUESTION 1C.5(c):

For n=3 and L = 1.0 m, the wavefunction equation is

x (m)	Ψ (x) (m-1/2)
0
1/6 = 0.167
2/6 = 0.333
3/6 = 0.5
4/6 = 0.667
5/6 = 0.833
1

Then the plot of Ψ (x) vs. x gives:

According to the graph, at n=3 and L = 1 m, the wavefunction equation has two nodes

• 1st node => at x = 2/6 =  0.333
• 2nd node => at x = 4/6 = 0.667

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ANSWER - QUESTION 1C.5(d):

The number of nodes present in a wavefunction is realted to n. The number of nodes is

• For n = 2 => the wavefuntion has one node
• For n = 3 => the wavefunction have two nodes

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ANSWER - QUESTION 1C.5(e):

the plot of probability density distribution isgiven by Ψ(x)²:

For n = 2 and L = 1 m

x (m)	Ψ (x)2 (m-1)
0
1/4 = 0.25
2/4 = 0.5
3/4 = 0.75
1

Then the plot of Ψ (x)² vs. x gives:

When Ψ (x)² reachs the maximum value it is most likely to find the particle. In this case, this occurs at x = 1/4 = 0.25 and at x = 3/4 = 0.75

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ANSWER - QUESTION 1C.5(f):

For n = 3 and L = 1 m, the equation of probability density distribution is:

x (m)	Ψ (x)2 (m-1)
0
1/6 = 0.167
2/6 = 0.333
3/6 = 0.5
4/6 = 0.667
5/6 = 0.833
1

Then the plot of Ψ (x)² vs. x gives:

When Ψ (x)² reachs the maximum value it is most likely to find the particle. In this case, this occurs at x = 1/6 = 0.167, at x = 3/6 = 0.5 and at x = 5/6 = 0.833