The nuclide ¹⁸F decays by both electron capture and β⁺ decay. Find the difference in the energy released by these two processes. The atomic masses are ¹⁸F = 18.000950 and ¹⁸O = 17.9991598.

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hey everyone in this example we need to state with the difference in energy released by a sulfur 31 new glide that undergoes electron capture and beta decay. If the atomic mass of 31 sulfur is 30. AM use and 31 phosphorus is 30. AM us. So we are going to recall the following formula where we can calculate the mass defect or delta M. Which is equal to the sum of our mass of our reactant subtracted from the sum of the mass of our products. So beginning with our given isotope in the prompt we have sulfur 31 recall that sulfur on the periodic table has an atomic number being 16. So we can fill that in and it's undergoing beta decay. So to undergo beta decay, we would add a beta particle which has a mass number. We recall of zero and an atomic number of negative one. So we're going to add everything across. And what we would get Is now our second isotope mentioned which is sulfur 31 And it's going to have atomic number 15. So to calculate our change in mass effect, We would say that we have our massive our reactant. So our reactant are going to be our sulfur and our beta particle here. So we would have from the prompt a atomic mass of 30.979557 am use this is then we can actually not close off the prentice yet because we're going to add this to our second reactant which is going to be the atomic mass of our beta particle which we should recall is 0. AM use this is then subtracted from the atomic masses of our products. Which in this case we just have one product and above that is given to us as 30.973762 AM use So what we would get for our mast effect value here for this first decay is a value equal to 0. AM use. And now according to the prompt, we also go through a second beta decay here. So what we would have is just going to rewrite this reaction. So we'll show this in a different color. We have sulfur 31. It still has atomic number 16 and it's going to form a beta decay where we have our beta particle with a mass number of zero. We recall and an atomic number of plus one. Now since this is the reverse and this is going to be added to Our phosphorus 31 which also has atomic number 15. And now we want to calculate mass effect Delta M equal to the sum of our masses of our reactant. So what we would have so far is our only reactant which has a mass for sulfur 31 being 979557 am use This is then subtracted from the sum of our masses of our products. So for our product we have our beta particle which we recall has a mass of 0. which is then added to our mass for phosphorus 31 given in the prompt as 30.973762 A. M. U. S. And so what we would get here for our mass defect is a value equal to 0.0054264. So now we're going to take the two mast effects that we've calculated and subtract the higher one from the lower one. So our higher value we found was 0. Six am use which is then subtracted from our lower mass defect value being 0.005464. And sorry that is 0.00542. We'll just keep this in right actually. So 0.0054264 am us. So this gives us a difference equal to a value of zero 000917 AM use. And because we need to give the final answer and units of energy for our decay, we should recall that the units are going to be in mega electron volts. So we're going to convert from A. M. U. S. Two mega electron volts. So we're going to recall the following conversion factor where one a.m. U. Is equal to 931.5 mega electron volts. Now we're able to cancel out our units of um use and what we're going to get. And just to make a quick correction here for our value of the second mass defect. That should be 5 to 4. So we're just going to make these quick corrections here. And this would change this value here so that it's now 0.1097 AM us. We still have our ammu units cancel out. And when we multiply this value by 9 31.5 we're going to get our final answer here equal to 1.0 to 20 mega electron volts. So this value here would be our final answer. And sorry about that. This is going to be our final answer here for our difference in energy released by the sulfur 31 new Clyde undergoing electron capture and beta decay. So I hope that everything I reviewed was clear. But if you have any questions, just leave them down below and I will see everyone in the next practice video

The nuclide ¹⁸F decays by both electron capture and β⁺ decay. Find the difference in the energy released by these two processes. The atomic masses are ¹⁸F = 18.000950 and ¹⁸O = 17.9991598.

Verified Solution

Key Concepts

Nuclear Decay Processes

Energy Release in Nuclear Reactions

Atomic Mass and Mass Defect

The nuclide 18F decays by both electron capture and β+ decay. Find the difference in the energy released by these two processes. The atomic masses are 18F = 18.000950 and 18O = 17.9991598.

Verified Solution

Key Concepts

Nuclear Decay Processes

Energy Release in Nuclear Reactions

Atomic Mass and Mass Defect

The nuclide ¹⁸F decays by both electron capture and β⁺ decay. Find the difference in the energy released by these two processes. The atomic masses are ¹⁸F = 18.000950 and ¹⁸O = 17.9991598.