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Ch.20 - Radioactivity and Nuclear Chemistry

Chapter 20, Problem 68

Calculate the mass defect and nuclear binding energy per nucleon of each nuclide. a. Li-7 (atomic mass = 7.016003 amu)

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Hey everyone in this example We're told that for sodium 23 with an atomic mass of given here we have to figure out what its nuclear binding energy and mass effect per nucleus of each nuclei ID would be. So we should recall that mass effect can be found from taking our atomic number Z. Multiplied by our mass of Hydrogen one. Which is then added to our atomic mass a minus our atomic number Z. Multiplied by the mass of a neutron. So it has a mass number of one and an atomic number zero. And then just to make room for everything, we're going to move this over. This is lastly subtracted from our mass of our isotope. So to calculate our mast effect for sodium 23, we would say that the mast effect is going to equal our mass number for sodium which according to our periodic tables. Or sorry, our atomic number Z for sodium according to our periodic table is going to be 11. This is multiplied by the mass of hydrogen which we should recall for hydrogen one as an isotope is 1.783 AM us. This is then added to our mass number a. Which given in the prompt for our isotope is included in the name being 23. Subtracted from the atomic number of sodium, which we recall is equal to 11 according to our periodic table. And then this is multiplied by our massive our neutron which we should recall is 1.866 A. M. Use So continuing on to make more room. We have this then subtracted from our massive our isotope which in the prompt is given to us as 22. AM use. So what this is going to give us is a value equal to 0.20028 AM Use. So now that we have our mass defect value, this is going to be our first answer for the prompt which is our mast effect of sodium 23. And now we're going to continue on to part two which is finding the binding energy of sodium 23. So we're going to take our mass effect value above which we found at 0.200 to eight a.m. U's. And we're going to multiply by the conversion factor. Where we should recall that for one a.m. You we have 931. mega electron volts. Now because this is for the binding energy of sodium 23. We want to multiply our denominator in am use for 11 nucleons Which would be our atomic number for sodium being 11. So this is for one a.m. U. Of 11 nucleons of sodium 23. And what we would be able to do is cancel our units of AM use leaving us with units of mega electron volts per nucleons. And this would give us our answer for the binding energy equal to 16.96007455. We have units of mega electron volts per nucleons, so we can round this to about 16.960 mega electron volts per nucleons. So this here would be our final answer for the nuclear binding energy. So everything highlighted in yellow represents our two final answers. To complete this example. I hope that everything I reviewed was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.