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Ch.21 - Radioactivity & Nuclear Chemistry

Chapter 21, Problem 69

Calculate the quantity of energy produced per gram of U-235 (atomic mass = 235.043922 amu) for the neutron-induced fission of U-235 to form Xe-144 (atomic mass = 143.9385 amu) and Sr-90 (atomic mass = 89.907738 amu) (discussed in Problem 57).

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Hey everyone in this example, we need to figure out how much energy is created. Program of Nittany um to 37 from neutron induced fission of plutonium to 37 were given its atomic mass as well as its product being barium 1 40 its atomic mass and rubidium 95 its atomic mass. So we're going to recall that to calculate the energy. We want to calculate our mass defect delta. M. And we can find that by taking the sum of our mass of our reactant. Subtracted from the sum of our mass of our products And undergoing through neutron induced fission, we would react with one neutron. So it has mass number of one and an atomic number of zero. This produces our product Barium 1 40 rubidium 95. And we would also produce two moles of our neutron here. So continuing on our neutron induced fission, we have our Nittany um to 37 Still reacting with a neutron to produce barium Rubidium 95 and now two moles of our neutron. Now, what we should recognize is that we're going to cancel out one of these neutrons here with the two moles here. And this would give us for our final result. Newtonian to 37 produces barium 40 plus one mole of a neutron. So now we're going to calculate our mass defect delta M. We're going to take our massive our reactant which is kept in a um to 37. And according to the prompt a has a mass of 2 37.482 AM use. This is then subtracted from the mass of our products. Where our first product Barium 1 40 has a mass given in the prompt of 39.910605 AM Use added to the mass of our second product being rubidium 95 which has a mass of 94.929303 am use given in the prompt and then added to our massive are neutron which we should recall is equal to 1.866 A. M. Use so just to move things over We are going to simplify this so that we get our mast effect value equal to 1.1996 g. Our grams. So we're able to cancel out grams. We're left with kilograms for a mass effect. And this is going to give us a result equal to 1. times 10 to the negative third power kilograms as our mass defect value. And now that we have our value for mass defect, we want our final answer. Energy to be in jewels. So we're going to recall the following equation where we have energy equal to mass times the speed of light squared. So this is going to equal our mass which above we stated is 1.1996 times 10 to the negative third power kilograms which is then multiplied by our speed of light which we recall is 3.0 times 10 to the eighth Power and we have units of meters per second and this is going to be squared. So this is going to give us a result equal to 1.797 times 10 to the 14th Power. And we have units of kilograms times meters squared, divided by seconds squared. We should recall that one jewel is equivalent to one kg times meters squared, divided by seconds squared. And so we can say that our final answer for energy is that it's equal to 1.797 times 10 to the 14th power jewels. So this would be our final answer here As our energy created program of Napkin Ium to 37. So what's highlighted in yellow is our final answer. I hope that everything I viewed was clear. If you have any questions, just leave them down below and I will see everyone in the next practice video