Rutherford Gold Foil Oil Experiment Example 1

Jules Bruno
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in this example question. It says the gold foil Rutherford used in his experiment had a thickness of approximately 6.0 times 10 to the minus three millimeters. If a single gold Adam has a diameter of 2.9 times 10 to the negative eight centimeters. How Maney Adams thick was Rutherford's foiled. All right, so here we need to determine what is our end amount here. They want us to determine the amount of atoms so the number of atoms will be our end amount and are given amount is just the value that possesses only one unit connected to it. Within this question, we have to values being given here. They're saying that we have 6.0 times 10 to the minus three millimeters, and in this other value, it's actually not alone. It's actually a conversion factor. It's telling us one Adam has this value. So the conversion factor is one. Adam is 2.9 times 10 to the negative eight centimeters. Remember, we're gonna start with are given amount, which is just a value alone that is not connected. Toa another unit. So we have 6.0 times 10 to the minus three millimeters. That is our given amount. Remember, we use conversion factors to go from, are given amount, tow our end amount and realize here that if I can convert these millimeters into centimeters, they can cancel out with these centimeters. And in that way we can isolate Adams at the end. So that is the approach we're going to take. So our first conversion factor is just going to be converting millimeters to meters. Remember, the coefficient of one goes on the side with the metric prefix meaning one Millie is 10 to the negative. Three millimeters can't slop Now I have meters for conversion factor to now I'm gonna go from meters, two centimeters. One senti is 10 to the negative too. Now that I have centimeters, I can now bring in my conversion factor from the question itself which is conversion factor three So 2.9 times 10 to the negative. Eight centimeters per one. Adam. So here centimeters cancel out and I'll have Adams at the end. Make sure you put this in parentheses in your calculator. Otherwise you make a thing incorrect. Answer. And remember, we do that every time. We have times 10 20 power if you that correctly, you'll get as your final answer to one times 10 to the four acts, so we would have 2.1 times 10 to the four atoms in terms of another atoms involved in the thickness of Rutherford's foil.