# Homework Help: Solving for y-intercept on a wavefunction graph

1. Apr 28, 2010

### 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

2. Apr 28, 2010

### zachzach

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?

Last edited: Apr 29, 2010
3. Apr 29, 2010

### Linus Pauling

I'm lost. What equation describes teh 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...

4. Apr 29, 2010

### zachzach

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.

5. Apr 29, 2010

### zachzach

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

6. Apr 30, 2010

### 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?

7. Apr 30, 2010

### Linus Pauling

I am also asked (and am confused about) computing the probablity 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?

8. Apr 30, 2010

### zachzach

Yup.

9. Apr 30, 2010

### zachzach

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.