Get Down

And

Scatter!!!

 

Rutherford Scattering

By

Kate Perry

And

Scott Hesser

 

Abstract:

This lab simulates the Rutherford scattering method of investigating atomic nuclei. We shoot a cylindrical target with a ball bearing and measure the angle at which it bounces off the target. The majority of this paper is spent deriving the geometry we use to find the size of the target from our measured angle. The simulation of Rutherford’s method is rough, but the general concept and geometry are present.

Introduction:

In 1911 Earnest Rutherford proposed an experiment to probe the interior of an atom by bombarding it with alpha particles. Rutherford was able to determine the size of the nucleus by analyzing the scattering pattern of the alpha particles. This lab demonstrates how it was possible for Rutherford to determine the size of his target from the scattering pattern. Of course, alpha particles scatter off the nucleus because of similar electric charges, but in this lab we use actual physical collisions. The geometry is similar, but it is important to remember that there are other factors to consider when working at the atomic level.

 

Theory:

When Rutherford set up his original scattering experiment it looked similar to our set up shown in Figure 1. A stream of alpha particles came in from where our air gun is located and hit a target in the center of a round chamber. The targets were the positive charges located in a piece of gold foil (at this time scientist knew atoms possessed positive charges, they didn’t know they were clumped together in a tight nucleus). The alpha particles reacted when they hit the chamber walls and left a mark, just as our ball bearings do. Rutherford then measured the scattering angle to determine the size of the target.

In Rutherford’s experiment, the scattering angle was easy to measure. The nuclear target was so small that there was no difference between the scattering angle, q, and the arc f (Figure 1). However, in this experiment, the target has a significant size and therefore the scattering angle is not directly measurable. The significant size of the target also has another consequence. In order to accurately model the target with a particle gun, the gun must be fired several times as it is moved along the cross section of the target. To determine the size of the target, the distance along the cross section from which the particle was shot must be known. More importantly, this value, called the impact parameter, b, must have the center of the target as the reference point. Only the edge of the target cross section is known, so again there is another un-measurable value needed to determine the size of the target. The size of the particle can be determined from the impact parameter and the scattering angle. However, before this can be done, each must be found in terms of measurable parameters. The following will derive just that.

In this experiment, D is the displacement of the gun from an arbitrary position (refer to Fig. 1 for this discussion). We set D=0 as close as possible to the target. By shooting the target at intervals away from D initial we get a scattering picture. The target is round so we should get a symmetric picture. By finding the center of symmetry for this picture we can label a value of D as the center Do. Do becomes b=0. This b is the impact parameter or the distance along the cross section of the target from its center.

Rutherford assumed the mass of the nucleus was much greater than the mass of the alpha particle and therefore the nucleus did not move during the collision. This means the particle is reflected as if it hit a solid wall. This applies for our experiment. If we turn to Figure 2 now, we can see that when the particle hits the target we can draw a line tangent to the target near the point of contact. We can consider this line the flat wall the particle reflects off. We know that the angle of incidence equals the angle of reflection, or in the diagram b1=b2. Note that Rutherford, and as we will, treated this as the collision between two point sources. Because of this we want to draw that tangent line at a point not on the edge of the target, but through the center of the particle. Thus our tangent line is R+r distance from the center of the target. We take this into consideration in our next step.

Looking again at Figure 2 we can see that n=b1 because they are opposite angles of two intersecting lines (excuse my lack of technical terminology on the geometry, if you want the exact Euclidean terms, go ask Euclid). We can also see that b2+n=q. Because b1=b2 we can say that b1=b2=(1/2) q. Now if we look at Figure 3, which is essentially Figure 2 with some different notation, we want to prove that –2=b1=b2. Lets look at triangles x and y. Both are right triangles. Also n and 1 are complimentary. Now there is some rule that because of this, n=–2. This tells us that the angle formed by b and R+r is 1/2 of q so we can write cos(1/2)q=b/(R+r).


If we look at Figure 1 we can also see that sina=b/(R+r). This is fairly obvious. This means sina= cos(1/2)q. Now, using some trig identity (which we all know there are a billion of them and at some point in our career we will waste hours blindly looking around for one), we say:

cosa= sqr(1-cos2(1/2)q)=sin(1/2)q.

We’ll have to refer back to these values latter for some substitution, so keep them in mind.

Our next step is to show that the following expression is true:

(sin(f-q))/(R+r)=(sin(q–––+a))/S.

To prove this we turn to Figure 4, which is a small section of Figure 1. We’ve cut out a lot so we can focus in on the most relevant angles and distances. The law of sines tells us (referring back to Figure 4 while I go through this) that:

sin–e/(R+r)=sin–f/S

Now we need to show that –e=q-f and –f=q+a. First, we know a must be the other part of –e because the two are opposite angles formed by a ray (R+r) passing through two parallel rays (x and y) therefore –f=q+a. Now to prove –e=f-q. We take advantage of ray y. Now the triangle formed by sides T, part of S and part of y have angles e, q, and p-f. Now the sum of these angles is p so adding them gives us p-f+–e+q=p. Now cancel out p and do some simple algebra and we see that –e=f-q.

Now lets just restate what we have proved:

(sin(f-q))/(R+r)=(sin(q–––+a))/S.

Remember that the goal is to find R+r in terms of measurable parameters. The rest of the process is straightforward algebra and goes relatively quickly. From here we use the addition formula sin(q–––+a)= sinacosq+cosasinq, and a little bit of algebra we can see that:

sin(f-q)=(R+r)/S (sinacosq+cosasinq).

And next we substitute and change this to:

sin(f-q)=(R+r)/S (cos(1/2)qcosq+sin(1/2)qsinq).

From here we use another addition formula to show that

sin(f-q)=(R+r)/S cos(q-(1/2)q).

Which reduces down to:

sin(f-q)=(R+r)/S cos(1/2)q.

Now isn’t this fun? Hang on, we’re almost there. We have shown before that cos(1/2)q=b/(R+r). If we substitute this in to the right side of the equation we find that the (R+r)’s cancel out and we are left with b/S. So now we have:

sin(f-q)=b/S.

OK, one last step. If we do the arcsin function on both sides we get:

f-q=arcsin(b/S).

Then we do a little algebra and WAMO!, we have:

q=f-arcsin(b/S).

Now we have our un-measurable values in terms of measurable parameters. The next step is to relate them to R+r. Now if we plot on a graph all our values of b versus the corresponding values of cos (1/2)q we will get a linear pattern. The slope of the best-fit line for this data will be the value R+r. Does that make sense? Remember we said before that cos (1/2)q=b/(R+r), Therefore,

b=(R+r) cos (1/2)q.

 

Experimental Procedure:

The set up for this lab is relatively easy. A tub, with a centered target and gun mount come prepackaged. Line the inner edge of the tub, or chamber, walls with colored wax paper that will leave a mark when the bb strikes it. This is a good point to label the center point of your paper, this is the mark where a bb shot directly at the center of the target would hit if it passed through the target. Load the gun and move it to a position where the bb just grazes the target, this is D=0. Move the gun in one turn and shoot the target 10 times and label the scattering marks. Repeat this until you come to the other end of the target. This is a laborious part of the lab and 10 shots might be overkill, but we want a good grade so we throw in the extra effort. Then remove the paper and measure the distance from the center point to the marks and find each groups average. Then get your calculator.

 

Results and Discussion:

Once the average distances from the center mark to the impact points are found, these values should be divided by the radius of the tub, S in the Figures, to convert them into the angles f. Now we have a list of values of D and the corresponding values of f. See Table 1. We know want to put D in terms of b. We do this by plotting a chart of D versus f. See Plot 1. Notice that the data is symmetrical. This is because the target is a cylinder and once we began shooting past the center of it the balls reflected in a similar manner in the opposite direction. (At this point we could have assigned negative values for the reflections in the other direction, but this would have just added confusion and its a simple obstacle to work around.) We draw a best-fit curve around this data and find the point of symmetry. This value of D, which we call Do equals 3.15cm. This also equals b=0, and if we do a shift in the data we have a list of b and the corresponding values of f. See Table 2.

Next we use the equation, q=f-arcsin(b/S), to convert f into q. See Table 3. We then make another table of b versus cos (1/2)q. See table 4. By plotting the data of Table 4 we can draw a best fit line to find the slope and the value of R+r. At this point we see where it may have helped to have given directional signs of the angles. If we plot the positive values of b, or the right side of the target, we find that we have a positive slope. See Plot 2. If we examine the negative values of b, or the left side, we see that we have a negative slope. See Plot 3. This is not a big deal. Just work with the absolute value of the two slopes.

If we ask the computer to draw a best-fit line and give us the slope, we find that the right side has a slope of 2.93±.2 and the left side has a slope of 2.94±.2. In terms of R+r, these are values in cm., and we shall work with the average of the two, 2.935±.2 cm. Now the known radius of our bb, r, is .15cm, and therefore our value of R is 2.785±.2 cm. Now we measured the radius of a target and found it to be 2.8cm. If our ruler measurement is accurate that gives us .5% error value, however, the procedure truly is not that accurate and the margins of error are fairly wide.

 

 

 

 

Conclusion:

The results of this lab were extremely accurate, and from this we can deduce that Scott Hesser and Kate Perry take damn good data. The radius of the target was found to be 2.785cm, and it is difficult to put an error value on this because of the lack of precise measuring tools. Rutherford’s scattering method is a sound way of analyzing the size of a round target.

 

References:

Beiser, Arthur. Concepts of Modern Physics. New York: McGraw-Hill Book Company, Inc. 1963.