How much energy must be supplied to break a single aluminum-27 nucleus into separated protons and neutrons if an aluminum-27 atom has a mass of 26.9815386 amu? How much energy is required for 100.0 g of aluminum-27? (The mass of an electron is given on the inside back cover.)

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welcome back everyone. If a silicon 29 atom has a mass of 28.9 76 49 am us. What is the amount of energy needed to break a single Silicon 29 nucleus into separate protons and neutrons? What is the amount of energy required for 150 g of Silicon given the mass of our electron here? So what we first want to make note of is that from the prompt? We have the mass number of silicon being 29. And on our periodic table when we find silicon, we see that it relates to the atomic number being 14. Where we would recall that our atomic number characterized by the symbols E tells us our number of protons meaning we would have protons. And because this is a neutral atom of silicon, we also would have 14 electrons match the same number of our protons here. So we want to determine first our mass of our silicon 29 nucleus. And we should recall that this would be calculated from taking our mass of our Silicon 29. Let's say its atomic mass. Rather instead of complicating it, we have the atomic mass of our Silicon which we would subtract from its nuclear mass. And so to determine the mass of our silicon nucleus, we have the atomic mass of silicon 29. Given in the prompt as 28.9 76 47 am. Use subtracted from its nuclear mass which we can determine by taking its 14 electrons and multiplying it by the mass of the electron given in the prompt as 5.485799 times 10 to the negative fourth power am use I'm sorry this is 799 here. This is going to result in a value equal to 28.996 15 A. M. Use. Now that we know the mass of the silicon 29 nucleus, we need to calculate the mass defect of this silicon 29. So recall that for mass defect represented by delta M. This is mass defect. This is found from taking our mass of our protons plus our mass of our neutron Which is then subtracted from. And sorry, this is a parenthesis at the end here. So this sum is subtracted from the mass of our nucleus that we calculated above for Silicon 29. So, plugging in what we know, we would say that our mass defect is equal to for our massive protons. As we stated, silicon corresponds to atomic number 14. So we would take our 14 protons and multiply it by the mass of a proton. Which we should recall is equal to the value one point oh oh 72765 A. M. Use this is then added to our mass of our neutron. So we should have a bracket here actually. So now we have added to this are massive are neutron were to determine our number of neutrons. Since we know that the mass number of our silicon 29 is 2029 here recall that are mass number is calculated by taking our protons added to our neutrons. And so we would take our 29 mass number and subtract it from our 14 protons to get a total of 15 neutrons, meaning we can go back to our formula down here and take our 15 neutrons and multiply it by the mass of a neutron, which we should recall is one point oh 86649 A. M. Use. So this completes our brackets here where now we're going to subtract from our massive our nucleus which we determined above. And I'll underline it in pink here so that we can plug it in as 28.968815 a. M. Us for our silicon 29 nucleus. This is going to result in a mass defect equal to a value of zero point A. M. Use. Now that we have our mass defect of our Silicon 29. We want to calculate the energy change. So that would be calculated as delta E. Which we should recall is equal to our mass effect. Two times the speed of light squared. So plugging in what we know we would have our mass defect value, which we just determined above as 20. 3029618 86 A. M U S. Where we're going to multiply by a conversion factor to go from AM use two g. So we would recall that we have avocados number as six point oh 22 times 10 to the 23rd power AM use for one g. Where then we want to multiply by our next conversion factor to cancel out that gram unit to go from 10 to the third. Power, grams equivalent to one kg multiplying by our next conversion factor, we would plug in Our speed of light, which sorry, is not a conversion factor. So our speed of light term is going to be 2.99797, sorry 2. 79246 times 10 to the eighth Power with units of meters per second. And this is going to be squared. So canceling out our units, we're going to get rid of um us, we're going to get rid of grams. What we're left with is kilogram squared times meter squared times seconds squared. And we want to recall that kilogram squared times meter squared times or divided by seconds squared is equal to one jewel. So this is equivalent to one Juul, meaning that for our final result for our Energy change, we're going to have a value equal 3.9- 50129 times 10 to the negative 11th power. And we have units of jewels. And this is going to be per silicon 29 nucleus required. So this is the energy required per silicon 29 nucleus. This is going to be our first answer for our prompt Where our second part of our prompt asks us for this energy required. But for 150 g of silicon 29. And so We're going to use the same calculation where we would have our energy change now equal to 150g of silicon, which is multiplied by our molar mass of Silicon 29 which we should recall is given in the prompt as 28. g for one mole of Silicon 29. Now we want to multiply by our mass change for one mole of Silicon 29 which would be the same of our as our mass change for one mole of our silicon 29 nucleus. Which we determined to be 290.24 point 2630296186. And let's make this clear So . 0296186. And this is per mole of Silicon 29. Now continuing on our calculation, we're going to scroll down for more room. Our next brackets or parentheses rather is going to be our conversion factor to go from 10 to the third power grams to one kg where we will then multiply by our speed of light squared as we did before. So recall that we have our speed of light constant as 2.992 point 246 times 10 to the eighth. Power units of meters per second. And this is squared. So now canceling out our units, we can get rid of kilograms or not kilograms but rather we can get rid of grams because we have it in the denominator and numerator is here. We also can get rid of our units of moles and we're left with units of kilograms squared times meter squared times seconds squared since we have that square power there. So this is going to yield our energy required for 150 g of Silicon 29 equal to a value of 1. times 10 to the 14th power jewels. Where we want to convert to kill a jewels. Since this is per 150 g of silicon 29. So we're going to multiply by the conversion factor to go from one or 10 to the third power jewels equivalent to one kg jule, allowing us to cancel our units of jewels leaving us with kilo jewels. And this is going to result in our final answer Equal to 1. times 10 to the 11th power per 150 g of silicon 29. So this would be our second final answer as the energy required for 150 g of silicon 29. And sorry, that is below here. So I hope that everything I reviewed was clear. If you have any questions, leave them down below and I will see everyone in the next practice video.

How much energy must be supplied to break a single aluminum-27 nucleus into separated protons and neutrons if an aluminum-27 atom has a mass of 26.9815386 amu? How much energy is required for 100.0 g of aluminum-27? (The mass of an electron is given on the inside back cover.)

Verified Solution

Key Concepts

Mass-Energy Equivalence

Binding Energy

Molar Mass and Avogadro's Number