Solving for y-intercept on a wavefunction graph

[Linus Pauling]

1.

What is c?



2. 1 = integral of wavefunction² from -infinity to +infinity



3. From the graph, the area above the x-axis is 2/3 the total area. I solved the following integral (int) from -1 to +1:

2/3 = int(c²dx)

Obtaining 2/3 = 2c²

So c = sqrt(1/3) = 0.58 nm^-1/2

 

[zachzach]

1.

What is c?
2. 1 = integral of wavefunction² from -infinity to +infinity
3. From the graph, the area above the x-axis is 2/3 the total area. I solved the following integral (int) from -1 to +1:

2/3 = int(c²dx)Obtaining 2/3 = 2c²

So c = sqrt(1/3) = 0.58 nm^-1/2

Remember, your are integrating the wave function squared, so you cannot just look at THAT graph. You must realize that the graph to look at is that graph with every value squared. What happens to negative values and fractions?

 

[Linus Pauling]

I'm lost. What equation describes the shape of the wavefunction given in the picture? I Understand that I will square that function, then integrate between the boundaries setting the integral equal to one, then solving for c...

 

[zachzach]

I'm lost. What equation describes the shape of the wavefunction given in the picture? I Understand that I will square that function, then integrate between the boundaries setting the integral equal to one, then solving for c...

I'm not sure what function. I am just saying from the graph given, graph the square of the function (which you can just do visually by squaring all the $$\psi$$ values). Then you can pull off the integration.

 

[zachzach]

Also, you can just define the piecewise function. So -inf < x < -2 , psi = 0 and so on. Then just square each piece.

 

[Linus Pauling]

Ok, so the shape of the squared wavefunction is basically two smaller rectangles with height 0.25c^2 and base of length 1, and a bigger one with base of 2 and height c^2. Total area = the integral = 1 = (5/2)c^2, so c = 0.632

Correct?

 

[Linus Pauling]

I am also asked (and am confused about) computing the probability of finding thte particle between -1 and 1nm.

Do I simply take the integral/area of the "bigger rectangle" that goes from -1 to 1nm, whose are is 2c^2, then divide by the total area 2.5c^2, i.e. 80% probability?

 

[zachzach]

Ok, so the shape of the squared wavefunction is basically two smaller rectangles with height 0.25c^2 and base of length 1, and a bigger one with base of 2 and height c^2. Total area = the integral = 1 = (5/2)c^2, so c = 0.632

Correct?

Yup.

 

[zachzach]

I am also asked (and am confused about) computing the probability of finding thte particle between -1 and 1nm.

Do I simply take the integral/area of the "bigger rectangle" that goes from -1 to 1nm, whose are is 2c^2, then divide by the total area 2.5c^2, i.e. 80% probability?

Well now you know what c is and the probability of finding the particle between -1 and 1 is the integral of psi squared from -1 to 1.