2. Atoms & Elements
Rutherford Gold Foil Experiment
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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 3 millimeters. If a single gold atom has a diameter of 2.9 times 10 to the negative 8 centimeters, how many atoms thick was Rutherford's foil? Alright. 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 our given amount is just the value that possesses only 1 unit connected to it. Within this question, we have 2 values being given. Here, they're saying that we have 6.0 times 10 to the minus 3 millimeters. And in this other value, it's actually not alone. It's actually a conversion factor. It's telling us one atom has this value. So the conversion factor is 1 atom is 2.9x10-eight centimeters. Remember, we're going to start with our given amount, which is just a value alone that is not connected to another, unit. So we have 6.0x10-three millimeters. That is our given amount. Remember, we use conversion factors to go from our given amount to 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 atoms at the end. So that is the approach we're going to take. So our first converting factor is just going to be converting millimeters to meters. Remember, the coefficient of 1 goes on the side with the metric prefix, meaning 1 milli is 10 to the negative 3. Millimeters cancel out, now I have meters. For conversion factor 2 now, I'm gonna go from meters to centimeters. 1 centi is 10 to the negative 2. Now that I have centimeters, I can now bring in my conversion factor from the question itself, which is conversion factor 3. So 2.9 times 10 to the negative 8 centimeters per 1 atom. So here, centimeters cancel out, and I'll have atoms at the end. Make sure you put this in parenthesis in your calculator, otherwise you make it the incorrect answer. And remember, we do that every time we have times 10 20 power. If you do that correctly, you'll get as your final answer, 2.1 times 10 to the 4 atoms. So we would have 2.1 times 10 to the 4 atoms in terms of the other atoms involved in the thickness of Rutherford's foil.
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