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Ch.20 - Radioactivity and Nuclear Chemistry

Chapter 20, Problem 87

Find the binding energy in an atom of 3He, which has a mass of 3.016030 amu.

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hey everyone in this example, we're asked for an atom of beryllium nine with a mass of nine point oh 1 to 182 AM us. What is the binding energy. So to find the binding energy, we want to recall our formula for mass effect, which we should recall is represented by delta M. This is going to be set equal to our atomic number of our adam beryllium Z. Which is multiplied by the mass of hydrogen. Where we have an atomic mass number of one, an atomic number of one. This is then added to the difference between our mass number, which we recall is represented by the simple A subtracted from our atomic number which we recall is represented by the symbol Z. And then multiplied by the atomic mass of a neutron which has a mass number of one and atomic number zero. And then this is going to be subtracted from the mass of our isotope. So, plugging in what we know, we would say that delta M. Is equal to our atomic number of beryllium. We would recall to the periodic table and see that it has an atomic number equal to four, so Z is equal to four. And we would plug in four. We're going to multiply this by the massive hydrogen, which from our periodic tables will be a value of 1.783 AM. Use quickly the mass of hydrogen one. The prompt as nine. So we would have nine subtracted from our atomic number for beryllium which from the periodic table we would see is four. And then this is multiplied by our mass of our neutron. Which we should recall is a value of 1.00866 am use and then subtracted from the mass of our isotope. So I'm just going to make more room for that because that is given in the prompt as 24 or sorry it's equal to So we're going to take this entire portion here in our calculators. And this is going to give us a value of 0.62438 AM use. Now we want to recall that the binding energy should be in units of mega electron volts. So we're gonna multiplied by the following conversion factor to go from am use two mega electron volts. And we should recall that for one a.m. You we have 931.5 mega electron volts. Now we're able to cancel out am use we're left with mega electron volts as our final units. And what we're going to get is a value equal to 58.160997 mega. And sorry, that should be a capital m mega electron volts as our binding energy. So this here would be our final answer to complete this example. I hope that everything I explained was clear. If you have any questions, please leave them down below. And I will see everyone in the next practice video