WEBVTT physics/ap-physics-1-2/fullerton
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Hi everyone and welcome back to Educator.com.
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I am Dan Fullerton and in this mini-lesson we are going to go through Page 1 of the APlusPhysics worksheet on mass energy equivalents and you can find that worksheet by clicking on the link down below the video.
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Take a few minutes and see if you can work through that and then we will check the answers together.
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All right. Moving on to Number 1.
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If a deuterium nucleus has a mass of 1.53 × 10^-3 universal mass units less than its components, this mass represents an energy of?
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Well it must have become an energy, so that is going to be 1.53 × 10^-3 universal mass units and we want to convert that into mega-electron volts (MeV).
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I know that 1 atomic mass unit, universal mass unit is 931 MeV, so our units will cancel out there and I will come up with about 1.42 MeV -- Answer Number 2.
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Number 2 here -- the energy equivalent of 5 × 10^-3 kg is -- well to find the energy equivalent, E = mc² -- probably heard that one before.
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That is going to be 5 × 10^-3 kg × c (the speed of light in a vacuum), 3 × 10^8 m/s².
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Put all that into my calculator and I come up with an answer of right around 4.5 × 10^14 J -- Answer Number 3.
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On to Number 3 -- How much energy in mega-electron volts is produced when one-quarter of a universal mass unit of matter is completely converted into energy?
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Well, we have 0.25 universal mass units and we want to convert that into energy in MeV.
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I know that one universal mass unit gives you 931 MeV, so I can treat this as a unit conversion problem, so 0.25 × 931 = 233 MeV.
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Number 4 -- The energy equivalent of the rest mass of an electron is approximately -- well again, E = mc², where our mass, the mass of an electron is 9.11 × 10^-31 kg × c (the speed of light), 3 × 10^8 m/s²...
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...which implies then that the energy equivalent is going to be -- when I put all of that into my calculator -- about 8.2 × 10^-14 J -- Answer Number 2.
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Number 5 -- The energy produced by the complete conversion of 2 × 10^-5 kg of mass into energy is...
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Well E = mc² again, where our mass is 2 × 10^-5 kg, c (speed of light) 3 × 10^8 m/s² is going to give us an energy of right around 1.8 × 10^12 J, which is 1.8 tera-joules (TJ) -- Answer Number 1.
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Number 6 -- What is the minimum total energy released when an electron and its anti-particle, known as a positron, annihilate each other.
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Well, energy is mc² again, where our mass is going to be 2 times the mass of an electron because the anti-particle has the same mass, but an opposite charge.
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That will be 2 × 9.11 × 10^-31 kg and our mass times the speed of light in our vacuum, 3 × 10^8 m/s², which comes out to be about 1.64 × 10^-13 J -- Answer Number 1.
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Number 7 -- The energy required to separate the 3 protons and 4 neutrons in the nucleus of a lithium atom is 39.3 MeV.
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Determine the mass equivalent of this energy in universal mass units.
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We have 39.3 MeV and we want to find out its mass equivalent.
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Well we want MeV to go away and we want universal mass units and I know that 1 universal mass unit is equal to 931 MeV, so we are multiplying by 1 again.
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Our units make a ratio of 1 and I come up with, when I do this, 39.3/931 is right around 0.0422 universal mass units.
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Number 8 -- Which graph best represents the relationship between energy and mass when matter is converted into energy?
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Well, if E = mc² and our variables are (E) and (M) -- well C² is just a constant, so it looks like we have a linear relationship between mass and energy.
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That is going to be that direct relationship, Number 1.
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Number 9 -- The total conversion of 1 kg of the sun's mass into energy yields what?
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Well, let us use E = mc², where our mass is 1 kg, c (speed of light in a vacuum) 3 × 10^8 m/s² is just going to be 9 × 10^16 J -- Answer Number 4.
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That is a lot of energy for 1 kg of mass.
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Number 10 -- What total mass must be converted into energy to produce a gamma photon with an energy of 1.03 × 10^-13 J?
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Well, before trying to get that energy, we know E = mc² and we want the mass.
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Therefore mass is going to be equal to the energy divided by the speed of light squared or 1.03 × 10^-13/3 × 10^8 m/s² (speed of light in a vacuum)...
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...which gives us a mass equal to about 1.14 × 10^-30 kg -- Answer choice, Number 1.
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Two more here -- A tritium nucleus is formed by combining two neutrons and a proton.
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The mass of this nucleus is 9.106 × 10^-3 universal mass unit less than the combined mass of the particles from which it is formed.
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Approximately how much energy is released when this nucleus is formed?
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Well, to do this, let us convert 9.106 × 10^-3 universal mass units into MeV, where I know that 1 universal mass unit is 931 MeV.
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Universal mass units are going to make a ratio of 1 and I am going to end up with an answer of about 8.48 MeV -- Answer Number 3.
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Last one on this page -- After a uranium nucleus emits an α particle, the total mass of the new nucleus in the α particle is less than the mass of the original uranium nucleus.
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Explain what happens to the missing mass.
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Well, that missing mass must have been converted into energy.
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All right, if you struggled with this it would be a great time to go back and review the sections on nuclear physics.
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If this went well -- Terrific -- Now you are probably ready to go tackle some of the AP level problems.
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Thanks so much for your time everyone and make it a great day.