Nuclear Physics Revision

let's revise nuclear physics starting with a very famous equation the delta e is equal to delta m c squared this tells us if the mass changes by amount delta m there will be a corresponding energy that's been either released or absorbed which is delta e notice that there's a factor of c squared over here meaning that we're actually going to get a huge amount of energy for every kilogram that has been converted let's apply easy equal to mc squared to a simple beta decay equation we have carbon 14 which will decay into nitrogen 14 via beta minus decay meaning that we're going to have a beta minus particle which is an electron and then an anti-neutrino let's compare the masses before and after this nuclear reaction we can see that the mass the mass of the carbon before is this number here times 10 to power minus 26 and all the corresponding masses are given here let's figure out our change in mass so delta m let's say that this will be equal to m final take away m initial like that now our final mass will just be the mass on the right hand side over here so the mass of nitrogen 14 is just equal to 2.3252723 and then we also have an electron so that's going to be plus north point north node node nine one one and from that we're gonna need to take away the mass of the carbon 14 particle which is pretty similar so minus two point three two five three nine one four and just remember that all of these ants or all of these numbers are multiplied by 10 to the power of minus 26. now if we put this into a calculator we're going to actually get a negative answer which is minus 2.8 times 10 to the power of minus 31 kilograms so that's a very very small amount of mass that we have actually lost so this amount of mass will correspond to an amount of energy that in this case has been released delta e we can calculate that energy we can just take the absolute value in this case of the mass because this is about the change of the mass so all we need to do is just take this number 2.8 times 10 to the power of minus 31 and then multiply by c squared which is equal to 3.0 times 10 to the 8th squared and what we're going to find is that we get around 2.52 times 10 to the power of minus 14 joules and this may seem like a tiny number but this is the amount of energy that's been released from a single reaction which is actually very large onto electron positron annihilation we have a couple of those over here and they annihilate and they produce two gamma-ray photons to conserve momentum what we need to do is to determine the frequency of one of those photons so we're just going to use the fact that delta e is equal to delta m c squared meaning that delta e will be equal to delta m now my total change in mass is going to be twice the electron mass because remember the electron the positron have the same mass but opposite charge and they're fully converted to energy therefore delta m will be equal to 2 times 9.11 times 10 to the power of minus 31 then i'm gonna multiply by a factor of c squared which is 3.0 times 10 to the eighth don't forget the square now because i'm looking for the energy of just one of those photons and there's two of them i'm also going to just divide this whole expression by two meaning that this will essentially cancel out so delta e will then be equal to approximately 8.2 times 10 to the power of minus 14 joules so this is the energy that's been given to one of these photons so this one here will get eight point two times zero minus fourteen and this one here will also get temp eight point two times ten to the power of minus fourteen because we're looking for the frequency then all i'm going to do is just use that e is equal to hf meaning that the frequency will be the energy divided by planck's constant so it's going to be 8.2 times 10 to the power of minus 14 go divided by 6.63 times 10 to the power of minus 34 which is just going to give us around 1.24 times 10 to the power of 20 hertz remember if we're looking for the wavelength we will just use instead the good old equation that e is equal to hc over lambda on to some very very important definitions we're going to start off with binding energy so binding energy is the energy required to completely separate the nucleus into its constituents which are protons and neutrons you can think of binding energy as the energy required to separate the nucleus for instance you may have some protons and some neutrons they're quite happy together but they will require some energy to be inputted similarly you can kind of think of it almost as if you had let's say some sort of a particle which is in a hole in some sort of a energy well it will require some energy to escape this so this is an analogy to try and visualize binding energy now mass defect is the difference between the mass of a completely separated nucleus and the nucleus itself typically in our e is equal to m c squared equation this will serve as our delta m very often the questions will actually ask us to calculate the binding energy per nucleon which is essentially just the minimum energy required to remove a nucleon from the nucleus to calculate the binding energy we're going to use e is equal to delta m c squared where delta m is the mass defect and e is the binding energy and to find the binding energy per nucleon what we're going to do is we're going to take e is equal to m c squared or let's put in some deltas delta is equal to delta m c squared well once again delta m is the mass defect delta e is just a binding energy binding energy and then to get the binding energy per nucleon all we need to do is divided by n where n is the number of nucleons so number of nucleons so this quantity here that i've highlighted is the binding energy per nucleon so on to nuclear fusion and fission so fusion occurs when two light and nuclei combine to produce one heavy nucleus so here's one nucleus here's another one typically you need some very extreme conditions in order to overcome the coulomb electrostatic repulsion because they're both positive and once they're combined they can produce a larger heavier nucleus nuclear fission on the other hand occurs when a heavy nucleus is split into two smaller nuclei or two lighter nuclei a graph that we need to remember for the exam is uh given right over here this is really really important this is the binding energy per nucleon with respect to the nucleon number a notice a couple of highlights first of all the most stable element is iron which is element 56 over here right around this region actually of isotopes this is a pretty stable region over here elements to the right cannot yield energy through fusion as the binding energy per nucleon will decrease what does that mean this means that all of the elements on this side right here will yield energy only through fission rather than through fusion on the other hand elements to the left it's let's use a different color so elements to the left which are all of those elements over here cannot yield energy through fission as the binding energy per nucleon will decrease ultimately though fusion can release a lot more energy as the binding energy difference for lighter nuclei is far greater in other words this graph changes here considerably more it requires though on the other hand some pretty extreme conditions for fusion to occur including high temperature and pressure to overcome that coulomb repulsion so just to make it absolutely clear with induced nuclear fission the total binding energy will be greater because we can have an element over here which might decay into an element over here meaning that the binding energy has increased with fusion we also have an increase in binding energy because we may have an element here that is fused with an element here giving us a heavier element with a higher binding energy so in each of those cases the binding energy let's call it delta e let's pick a different color the binding energy will be increasing we can calculate that binding energy using delta e is equal to delta m c squared binding energy per nucleon example calculate the binding energy per nucleon of beryllium eight four we're given the mass of the nucleus is 1.329 times 10 power minus 26 kilograms the proton has a certain mass and the neutron has the mass given as well okay well first off we need to figure out the mass defect our mass defect delta m is equal to the mass of the all of these separated nuclei so let's say m2 take away m1 which is just the mass of the nucleus so in a way initially we have a whole bunch of particles together so this here is our before and our after is a whole bunch of particles which are fully separated quite far from one another okay well our mass of the fully separated particles will just be equal to 4 times the mass of a proton plus 4 times the mass of a neutron how do i know this well if we look over here we're given that it's beryllium for beryllium eight four meaning that we have four protons and eight nucleons all together meaning that we're going to have four neutrons as well so in order to calculate these all i need to do is get four times the mass of a proton which is 1.67 times 10 to the power of -27 add 4 times the mass of a neutron which is 4.0 times 1.67 times 10 to a power of minus 27 and from that what we need to do is take away our original mass which is the mass of the nucleus beforehand which is just this mass over here 1.329 [Music] times 10 to the power of minus 26. this will of course give me approximately 1.020 times 10 to the power of minus 28 kilograms and this is my mass defect our next step is to figure out our binding energy and this will just be equal to delta m c squared so this is going to be 1.020 times 10 to the power of minus 28 multiplied by a factor of c squared which is 3.0 times 10 to the power of 8 squared which is going to give us approximately 9.18 multiplied by 10 to the power of minus 12 joules because in this question we're actually looking for the binding energy per nucleon as it's being acts asked over here what we need to do is divide by the nucleon number the nucleon number in this case is just eight there are four protons and four neutrons giving us a total of eight nucleons so what i'm going to do is just divide my answer for the binding energy by the number of nucleons so i'm going to find that e over 8 is equal to just 9.18 times 10 to the power of minus 12 divided by 8. let's put this into a calculator and we're going to get approximately 1.148 times 10 to the power of minus 12 joules per nucleon now let's talk about induced nuclear fission chain reaction we have a single thermal neutron which is actually often moving quite slowly is absorbed by a nucleus and thus creating a heavier unstable nucleus which undergoes fission and it splits into two nuclei and thus more neutrons the process will then repeat itself so this neutron will be absorbed by this uranium nucleus which will then split into two and so on and so forth creating a chain reaction now let's revise the components of a nuclear reactor first off we have the fuel rods and they contain uranium fuel it's very important when you're answering questions in an exam to quote that we are talking about uranium fuel you will not get the marks if you just say that they contain the fuel next up we have the control rods they absorb some of the neutrons to control the rate of the nuclear reaction now what do i mean by that they're actually in certain weight such that one neutron from a previous reaction on average causes further fission this makes sure that the power output is constant so she'll just add in the word power that saying that we mean that we're talking about the power output we also have the moderator which is really important it slows down the fast moving neutrons to create thermal neutrons and on average they actually create the biggest probability of nuclear fission this way but all we need to write an exam is that it slows down the fast moving neutrons create thermal neutrons water is actually one example of a moderator which works extremely well and it is found to be used as a moderator in most nuclear reactors finally let's talk about nuclear fusion and temperature you need some extreme conditions to achieve nuclear fusion for instance the condition inside the core of a star you need some very high temperature and additionally you need some very high pressure in order to overcome the electrostatic repulsion between the nuclei remember the nuclei are positive so for instance if we have a few nuclei which are trying to fuse they're going to be experiencing some extreme electrostatic repulsion forces because they're both positive and two charges of the same charge repel at higher temperature there's actually a higher probability of nuclear fusion and this has come up a couple of times in an exam question for instance here's a question from ocr from 2004 so we have a figure which shows the probability of fusion with temperature t between deuterium and tritium and do deuterium and helium now tritium is just a form of hydrogen an isotope of hydrogen with one proton and two neutrons deuterium and helium on the other hand seems to have a lower probability suggest why the probability of reaction at a given temperature is smaller for deuterium and helium well helium has more protons and hence a greater charge remember tritium is just hydrogen so it only has one proton where as helium is well element number two so it's going to have two protons so it's going to be a lot more electrostatic repulsion that needs to be overcome therefore the well the electrostatic repulsion force between the deuterium and the helium is greater and we have scored full marks well done guys now that you've revised this your next step is to have a look at part one which is all about the nuclear atom and the strong nuclear force and beta plus and beta minus decay and have a look at this revision video over here to help you guys revise thank you very much for watching i'll see you in the next video

let&#39;s revise nuclear physics starting with a very famous equation the delta e is equal to delta m c squared this tells us if the mass changes by amount delta m there will be a corresponding energy that&#39;s been either released or absorbed which is delta e notice that there&#39;s a factor of c squared over here meaning that we&#39;re actually going to get a huge amount of energy for every kilogram that has been converted let&#39;s apply easy equal to mc squared to a simple beta decay equation we have carbon 14 which will decay into nitrogen 14 via beta minus decay meaning that we&#39;re going to have a beta minus particle which is an electron and then an anti-neutrino let&#39;s compare the masses before and after this nuclear reaction we can see that the mass the mass of the carbon before is this number here times 10 to power minus 26 and all the corresponding masses are given here let&#39;s figure out our change in mass so delta m let&#39;s say that this will be equal to m final take away m initial like that now our final mass will just be the mass on the right hand side over here so the mass of nitrogen 14 is just equal to 2.3252723 and then we also have an electron so that&#39;s going to be plus north point north node node nine one one and from that we&#39;re gonna need to take away the mass of the carbon 14 particle which is pretty similar so minus two point three two five three nine one four and just remember that all of these ants or all of these numbers are multiplied by 10 to the power of minus 26. now if we put this into a calculator we&#39;re going to actually get a negative answer which is minus 2.8 times 10 to the power of minus 31 kilograms so that&#39;s a very very small amount of mass that we have actually lost so this amount of mass will correspond to an amount of energy that in this case has been released delta e we can calculate that energy we can just take the absolute value in this case of the mass because this is about the change of the mass so all we need to do is just take this number 2.8 times 10 to the power of minus 31 and then multiply by c squared which is equal to 3.0 times 10 to the 8th squared and what we&#39;re going to find is that we get around 2.52 times 10 to the power of minus 14 joules and this may seem like a tiny number but this is the amount of energy that&#39;s been released from a single reaction which is actually very large onto electron positron annihilation we have a couple of those over here and they annihilate and they produce two gamma-ray photons to conserve momentum what we need to do is to determine the frequency of one of those photons so we&#39;re just going to use the fact that delta e is equal to delta m c squared meaning that delta e will be equal to delta m now my total change in mass is going to be twice the electron mass because remember the electron the positron have the same mass but opposite charge and they&#39;re fully converted to energy therefore delta m will be equal to 2 times 9.11 times 10 to the power of minus 31 then i&#39;m gonna multiply by a factor of c squared which is 3.0 times 10 to the eighth don&#39;t forget the square now because i&#39;m looking for the energy of just one of those photons and there&#39;s two of them i&#39;m also going to just divide this whole expression by two meaning that this will essentially cancel out so delta e will then be equal to approximately 8.2 times 10 to the power of minus 14 joules so this is the energy that&#39;s been given to one of these photons so this one here will get eight point two times zero minus fourteen and this one here will also get temp eight point two times ten to the power of minus fourteen because we&#39;re looking for the frequency then all i&#39;m going to do is just use that e is equal to hf meaning that the frequency will be the energy divided by planck&#39;s constant so it&#39;s going to be 8.2 times 10 to the power of minus 14 go divided by 6.63 times 10 to the power of minus 34 which is just going to give us around 1.24 times 10 to the power of 20 hertz remember if we&#39;re looking for the wavelength we will just use instead the good old equation that e is equal to hc over lambda on to some very very important definitions we&#39;re going to start off with binding energy so binding energy is the energy required to completely separate the nucleus into its constituents which are protons and neutrons you can think of binding energy as the energy required to separate the nucleus for instance you may have some protons and some neutrons they&#39;re quite happy together but they will require some energy to be inputted similarly you can kind of think of it almost as if you had let&#39;s say some sort of a particle which is in a hole in some sort of a energy well it will require some energy to escape this so this is an analogy to try and visualize binding energy now mass defect is the difference between the mass of a completely separated nucleus and the nucleus itself typically in our e is equal to m c squared equation this will serve as our delta m very often the questions will actually ask us to calculate the binding energy per nucleon which is essentially just the minimum energy required to remove a nucleon from the nucleus to calculate the binding energy we&#39;re going to use e is equal to delta m c squared where delta m is the mass defect and e is the binding energy and to find the binding energy per nucleon what we&#39;re going to do is we&#39;re going to take e is equal to m c squared or let&#39;s put in some deltas delta is equal to delta m c squared well once again delta m is the mass defect delta e is just a binding energy binding energy and then to get the binding energy per nucleon all we need to do is divided by n where n is the number of nucleons so number of nucleons so this quantity here that i&#39;ve highlighted is the binding energy per nucleon so on to nuclear fusion and fission so fusion occurs when two light and nuclei combine to produce one heavy nucleus so here&#39;s one nucleus here&#39;s another one typically you need some very extreme conditions in order to overcome the coulomb electrostatic repulsion because they&#39;re both positive and once they&#39;re combined they can produce a larger heavier nucleus nuclear fission on the other hand occurs when a heavy nucleus is split into two smaller nuclei or two lighter nuclei a graph that we need to remember for the exam is uh given right over here this is really really important this is the binding energy per nucleon with respect to the nucleon number a notice a couple of highlights first of all the most stable element is iron which is element 56 over here right around this region actually of isotopes this is a pretty stable region over here elements to the right cannot yield energy through fusion as the binding energy per nucleon will decrease what does that mean this means that all of the elements on this side right here will yield energy only through fission rather than through fusion on the other hand elements to the left it&#39;s let&#39;s use a different color so elements to the left which are all of those elements over here cannot yield energy through fission as the binding energy per nucleon will decrease ultimately though fusion can release a lot more energy as the binding energy difference for lighter nuclei is far greater in other words this graph changes here considerably more it requires though on the other hand some pretty extreme conditions for fusion to occur including high temperature and pressure to overcome that coulomb repulsion so just to make it absolutely clear with induced nuclear fission the total binding energy will be greater because we can have an element over here which might decay into an element over here meaning that the binding energy has increased with fusion we also have an increase in binding energy because we may have an element here that is fused with an element here giving us a heavier element with a higher binding energy so in each of those cases the binding energy let&#39;s call it delta e let&#39;s pick a different color the binding energy will be increasing we can calculate that binding energy using delta e is equal to delta m c squared binding energy per nucleon example calculate the binding energy per nucleon of beryllium eight four we&#39;re given the mass of the nucleus is 1.329 times 10 power minus 26 kilograms the proton has a certain mass and the neutron has the mass given as well okay well first off we need to figure out the mass defect our mass defect delta m is equal to the mass of the all of these separated nuclei so let&#39;s say m2 take away m1 which is just the mass of the nucleus so in a way initially we have a whole bunch of particles together so this here is our before and our after is a whole bunch of particles which are fully separated quite far from one another okay well our mass of the fully separated particles will just be equal to 4 times the mass of a proton plus 4 times the mass of a neutron how do i know this well if we look over here we&#39;re given that it&#39;s beryllium for beryllium eight four meaning that we have four protons and eight nucleons all together meaning that we&#39;re going to have four neutrons as well so in order to calculate these all i need to do is get four times the mass of a proton which is 1.67 times 10 to the power of -27 add 4 times the mass of a neutron which is 4.0 times 1.67 times 10 to a power of minus 27 and from that what we need to do is take away our original mass which is the mass of the nucleus beforehand which is just this mass over here 1.329 [Music] times 10 to the power of minus 26. this will of course give me approximately 1.020 times 10 to the power of minus 28 kilograms and this is my mass defect our next step is to figure out our binding energy and this will just be equal to delta m c squared so this is going to be 1.020 times 10 to the power of minus 28 multiplied by a factor of c squared which is 3.0 times 10 to the power of 8 squared which is going to give us approximately 9.18 multiplied by 10 to the power of minus 12 joules because in this question we&#39;re actually looking for the binding energy per nucleon as it&#39;s being acts asked over here what we need to do is divide by the nucleon number the nucleon number in this case is just eight there are four protons and four neutrons giving us a total of eight nucleons so what i&#39;m going to do is just divide my answer for the binding energy by the number of nucleons so i&#39;m going to find that e over 8 is equal to just 9.18 times 10 to the power of minus 12 divided by 8. let&#39;s put this into a calculator and we&#39;re going to get approximately 1.148 times 10 to the power of minus 12 joules per nucleon now let&#39;s talk about induced nuclear fission chain reaction we have a single thermal neutron which is actually often moving quite slowly is absorbed by a nucleus and thus creating a heavier unstable nucleus which undergoes fission and it splits into two nuclei and thus more neutrons the process will then repeat itself so this neutron will be absorbed by this uranium nucleus which will then split into two and so on and so forth creating a chain reaction now let&#39;s revise the components of a nuclear reactor first off we have the fuel rods and they contain uranium fuel it&#39;s very important when you&#39;re answering questions in an exam to quote that we are talking about uranium fuel you will not get the marks if you just say that they contain the fuel next up we have the control rods they absorb some of the neutrons to control the rate of the nuclear reaction now what do i mean by that they&#39;re actually in certain weight such that one neutron from a previous reaction on average causes further fission this makes sure that the power output is constant so she&#39;ll just add in the word power that saying that we mean that we&#39;re talking about the power output we also have the moderator which is really important it slows down the fast moving neutrons to create thermal neutrons and on average they actually create the biggest probability of nuclear fission this way but all we need to write an exam is that it slows down the fast moving neutrons create thermal neutrons water is actually one example of a moderator which works extremely well and it is found to be used as a moderator in most nuclear reactors finally let&#39;s talk about nuclear fusion and temperature you need some extreme conditions to achieve nuclear fusion for instance the condition inside the core of a star you need some very high temperature and additionally you need some very high pressure in order to overcome the electrostatic repulsion between the nuclei remember the nuclei are positive so for instance if we have a few nuclei which are trying to fuse they&#39;re going to be experiencing some extreme electrostatic repulsion forces because they&#39;re both positive and two charges of the same charge repel at higher temperature there&#39;s actually a higher probability of nuclear fusion and this has come up a couple of times in an exam question for instance here&#39;s a question from ocr from 2004 so we have a figure which shows the probability of fusion with temperature t between deuterium and tritium and do deuterium and helium now tritium is just a form of hydrogen an isotope of hydrogen with one proton and two neutrons deuterium and helium on the other hand seems to have a lower probability suggest why the probability of reaction at a given temperature is smaller for deuterium and helium well helium has more protons and hence a greater charge remember tritium is just hydrogen so it only has one proton where as helium is well element number two so it&#39;s going to have two protons so it&#39;s going to be a lot more electrostatic repulsion that needs to be overcome therefore the well the electrostatic repulsion force between the deuterium and the helium is greater and we have scored full marks well done guys now that you&#39;ve revised this your next step is to have a look at part one which is all about the nuclear atom and the strong nuclear force and beta plus and beta minus decay and have a look at this revision video over here to help you guys revise thank you very much for watching i&#39;ll see you in the next video

Transcript for:Nuclear Physics Revision

Transcript for:
Nuclear Physics Revision