What fraction of the a particles in Rutherford's gold foil experi...

Question

What fraction of the a particles in Rutherford's gold foil experiment
are scattered at large angles? Assume the gold foil
is two layers thick, as shown in Figure 2.9, and that the approximate
diameters of a gold atom and its nucleus are 270
pm and 1.0 * 10–2 pm, respectively. Hint: Calculate the cross
sectional area occupied by the nucleus as a fraction of that
occupied by the atom. Assume that the gold nuclei in each
layer are offset from each other.

Accepted Answer

welcome back everyone in this example, we have the approximate diameter of a gold atom being 270 PK meters, its nucleus has a diameter of 1.2 times 10 to the negative second power meters. And we need to calculate the fraction of the alpha particles in Rutherford's gold foil experiment that are scattered at large angles by using the cross sectional area occupied by the nucleus as a fraction of that occupied by the atom. We're assuming that the gold foil is four layers thick and the nuclei is equally offset from each other so that no cross section overlaps. So let's go ahead and draw sketch before we begin problem solving. And we should recall that are set up for Rutherford's gold foil experiment is we have an alpha particle source. So that's going to be this source here. So we have our alpha particle source and we're shooting our alpha particles towards our gold foil here. So to sketch our gold foil, we'll just use a highlighter and we're told that it's four layers thick. So we'll just draw four lines. So this is our gold foil. And recall that in Rutherford's experiment, we have a detector. So we'll use the color purple a detector which surrounded the experiment. And what we should recall is that as our alpha particles were being shot towards the gold foil, A majority of these alpha particles were aimed towards our detector here, however. And let's go ahead and actually add to our drawing for the gold foil. So we want to draw in our nuclei of our gold foil atoms. So we have our atoms and in the center is their nuclei. And what we should recognize is that although we see our alpha particles majority hitting the detector in this experiment, there were a few alpha particles that actually made their way towards the gold foil and were deflected by the nuclei of the gold foil. And so they were deflected by the detector in this direction. Whereas most of the alpha particles went forward straight across and hit our detector. And this is what this prompt wants us to solve is the fraction of alpha particles that are being deflected by the nuclei of our gold foil here. And recall that the point of Rutherford's experiment is that he discovered that atoms are mostly dense and empty space, but they have a positively charged nucleus, which is why they're being deflected by the gold foil, atoms, nuclei. So moving into our solution, we want to begin by finding the fraction of alpha particles that hit the nuclei per atom per layer of gold foil. So we would say that the fraction is equal to the area of our nuclei of our gold foil, atoms divided by the area of the atom of those gold foiled nuclei. And so we're going to plug in what we know and we also want to recognize that we need to multiply this by four because as the prompt states are gold foil is four layers thick. We also want to recognize that the nuclei of our gold foil as well as these atoms of gold are sorry, are going to be spierce slash circles. And so when we use our formula for area of our nuclei and area of our atom, we want to recall our formula for the area of a circle which is going to be, the area is equal to pi r squared. Now what we should recognize is that are in this formula is radius and the prompt gives us diameters for the atom and PK meters and for the nuclei also in PK meters at these two values. And so we need to go ahead and divide those values by two. So going into our formula, we would say that the fraction of our alpha particles that are deflected is equal to first beginning with our numerator, we have the area of our nuclei which is actually given us a diameter. In the prompt being one point oh times 10 to the negative second power PICO meters. We want to divide this by two. So we're going to place this in parentheses because according to our area formula for nuclei, which is a circle or sphere inside of our atom, we have to multiply this by pi and square this value. And then outside of our entire quotient here, we're multiplying everything by four. So we're multiplying by four and then in our denominator we want to plug in the area of our atoms. So according to the prompt, we have a diameter of our atom being 270 PK meters. So we want to also divide this by two. Since we wanted to be a radius value, we're going to multiply this by pi according to our formula for area of a sphere, which is also the shape of a gold atom. And this is going to be squared. And so in our calculators, we're going to carefully plug this in with our parentheses. And this is going to yield us a value of our fraction of alpha particles that are deflected equal to 5.5 times 10 to the negative ninth power. And our units of PICO meters are going to cancel out because we're dividing. And so this would actually complete this example as our final answer because now we know the fraction of our alpha particles that are deflected. And so this would be our final answer. So I hope that everything I went through is clear. If you have any questions, leave them down below and I'll see everyone in the next practice video

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