 so I have two terms now in neutron this fluctuation term is zero for x-rays so if I consider some scattering length for x-rays it is same everywhere and there is no fluctuation it is zero so there is no incorrect scattering but in case of neutrons it is not so and now I will go ahead and show you how to do it so there is no angle dependence for this so now we can experimentally find out the scattering length of an isotope which has got a concentration CK then the averaging depends on the concentration and it is simply CK BK if BK is the scattering cross section of one isotope it is concentration CK that is sum over CK BK sum over the entire concentration and similarly B average B square average nothing but if the BK is the scattering length for an isotope with concentration of CK then its square is the square average of the square so let me just write it down so if B average is nothing but concentration CK of an isotope and its scattering length BK and B average B square average is equal to same sum on the square of that square of that and this I can calculate the diffraction from this and also I will talk about the spin now let me write it down let me derive it in case of spin suppose a nucleus has a spin i and the neutron has a spin half so we know from quantum mechanics twice s plus one at the possible states actually in our bachelors we know we found out the mj values by this so now you have twice the nuclear spin is either parallel to neutron or antiparallel plus one so when you take plus one there is two into one plus one it is twice i plus two and when minus half it is twice i twice i so total is four i plus two possibilities so when it is nuclear and neutron spin are parallel then you have i plus half in the possible states spin states are twice i plus two and when they are antiparallel in the possible state twice i and these are the weight factors because when they are parallel let us consider the scattering length is B plus and when they are antiparallel let us call it B minus then i have B average for spin will be equal to now weightage for the if you see this total number of states are four i plus two the weightage for the plus means spin and nuclear spin parallel is twice i plus two so it will be twice i plus two upon four i plus two with respect to the total weight this is the weightage and when the scattering length is plus plus when the antiparallel it is twice i twice i divided by four i plus two and then the scattering length B minus do not ask me the question how we find out B plus and B minus yes we have to do a very elaborate experiment we have to make the neutron polarization parallel to the polarization of the nuclear nucleus and then we have to do it but right now theoretically this is the case then you have the averaging over the spin is given by when the nuclear spin is i very simple algebra now we have the case where you have the isotope and the spins so if I consider an isotope kth isotope with spin ik please excuse me because there is a capital and small letter but this is the same as the capital i that I was writing in the blackboard so we take the concentration ck of the kth isotope with with nuclear spin ik then ik plus one upon twice ik plus one into bk plus plus ik upon twice ik plus one bk minus so this is the expression for B average and same way the B square average is that only I put square of these terms and weightage are the same so now B average equal to let me write explicitly concentration of the spin with the isotope with concentration ck spin ik ik plus one this is for B average and same way for B square average just put dots there so I can calculate B average and B square average for an distribution of isotopes and spins with kth isotope having concentration ck and spin ik so this is the expression adding the isotope and spin inner formalism now for example I have just written down here the B plus for the most common isotope which is hydrogen hydrogen has a B plus of 1.04 into 10 to the power minus 14 and B minus minus 4.74 into 10 to the power minus 14 I requested to calculate the coherent and incoherent scattering cross section for proton using these values let me just do a few lines for you so that you become familiar with this so it is a simplest calculation because proton has a spin half neutron has a spin half so twice I plus 1 is free and twice I so the spin weightages are 3 and 1 so I am telling you to do the spin average scattering cross section so what you need to do is actually now 3 fourths into B plus plus 1 fourth into B minus which is equal to 3 fourths 1.04 into 10 to the power minus 14 meters and other one is minus 4.74 there is minus 1 fourth 4.74 into 10 to the power minus 14 meters now you can see that this is subtracting from each other question comes how come a scattering length is negative right now I will request you to accept it in case of neutrons it is possible to have positive and negative scattering length because of certain nuclear constraints if possible I will try to explain it later otherwise I will request you to accept this and because of this you will find sigma coherent which is equal to 4 pi into B average square in this case it will come out to be something like 6.42 barbs 1 burn equal to 10 to the power minus 24 centimeters square and the sigma incoherent equal to 4 pi which is you can have to also evaluate B square average in the same way I did it and you will find it will be coming around 80 barbs so now you see so sigma incoherent for hydrogen is much much larger than sigma coherent so what it means that means hydrogen hydrogen is a very strong scatter hydrogen is a strong scatter no doubt about it is a very strong scatter but it scatters incoherently so for a diffraction experiment hydrogen is not so suitable so I will just show you with this coherent and incoherent part the coherent part as I am repeatedly telling you it has got q dot rl minus rl prime and it has if you do an experiment if the material has large coherent cross section then you will find this coherent peaks but incoherent one which is 4 pi B square average minus B average square does not have an angle dependence it gives an incoherent background because of this if you use hydrogenous material you will find that in your experiment you have got a very large background and possibly the coherent peaks are just about rising on top of them so this is a disadvantage of doing diffraction experiments with hydrogen it does not mean hydrogen is not used hydrogen is also used in some of the scattering experiments and also some of you may be aware that H2O is a very strong moderator because it is very strongly scatters neutrons and thermal neutrons and can cause very efficient thermal thermalization of neutrons because it has got a very large cross section overall which is around 81 81 pounds I stop the module where I have introduced you to the coherent and incoherent scattering cross section for neutrons in the next part I will show you a comparison between neutrons and X-rays because X-rays are are most commonly used microscopic probe almost all of us and then I will stop today