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Exploring the Inside of Atoms through Collisions

Discover the fascinating world of atoms and their components by studying their collisions. Learn about the properties and behavior of particles through experiments with coins and delve into the realms of subatomic physics.

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Exploring the Inside of Atoms through Collisions

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  1. Experimenty v částicové fyzice Jiří Dolejší, Olga Kotrbová, Charles University Prague We have “standard ways” to discover properties of things around us and to look to the “inside” of objects (like the small alarm clock on the photograph). But these methods may not be applicable to atom and its components … we do not have any sufficiently small screwdriver, we even cannot look at it with sufficient resolution. With the best current microscopes we can only see the individual atoms like on this picture from the tunneling microscope, but not the inside of them.

  2. But already at the beginning of the 20th century E. Rutherford and his collaborators developed a novel method to study the inside of atoms. They shot a-particles towards a thin gold foil and discovered that the idea best corresponding to the experimental data is the idea of atom being almost empty with small heavy nucleus and electrons flying around. Today gold nuclei are collided at RHIC (Relativistic Heavy Ion Collider) today and we expect to learn more about the matter at collision. How methods based on collisons can work? Let us try! A first attempt: I will try to use collision instead of a screwdriver: I succeeded in getting inside! But not sufficiently deep into the structure. Maybe I need higher energy… Probably it would be better to start with a simpler object…

  3. Atoms, nuclei, particles … Implicitly we look at them as small balls. Instead of playing with balls I suggest to flick coins and study their collisions. Put one coin on a smooth surface, flick another coin against it, observe the result. Disclaimer: The collisions of euro and dollar have neither political nor economical meaning. After playing a while with different coins you will get some experience. Look at the following situations and decide, which of the coins is heavier (arrows show velocities). B C A

  4. I hope you answers are correct: both coins have the same mass in B, blue one is lighter in A and heavier in C. You can repeat the experiment with “coins” and “doublecoins” glued together with doublesided tape: What about describing the scattering of coins in a manner usual in mechanics? Relevant variables are the masses m1, m2, velocities before and after collision v1 , v2 and v1', v2’. Important variables are energyE and momentum p. v1' m1 v1 Both energy and momentum in an isolated system are conserved: a b The red coin is at rest at the beginning m2 v2' v1= 0

  5. Let us work only with momenta: energy conservation... … and conservation of both momentum components Let us play with momenta: We would like to get rid of p2 and b and it is appealing to use for that purpose the relation We just rewrite the equations, make a square of each one and sum them: we can insert into energy equation:

  6. The last equation is a the quadratic equation for The discriminant is … and solutions

  7. Let us remind that we are looking for real nonnegative solutions. The simplest case is the case of equal massesm1 = m2: So in collision of two coins with equal masses the coin cannot be scattered backwards. It either continues forwards, maybe deflected, or stops. We can see from the conservation equations that if the coin stops, the target coin takes over the whole energy and momentum. Try it on a billiard!!! Instead of continuing the detailed discussion, we will plot the depen-dence of momentum after collision on the scattering angle: 1…projectile 2…target Only lighter particle can be scattered backwards The deflection of the heavier particle is limited

  8. Although we have played with coins, we used only the most general mechanical laws. Our results holds for collisions of any objects, including subatomic particles. The only “complication” is related to the pleasant fact, that particles can be accelerated to speeds close to the speed of light. The effects well described or predicted by the special theory of relativity appear - particles have bigger mass, unstable particles live longer. We can quite easily modify our calculation: Instead of counting kinetic energy we should deal with total energy we will insert into energy conservation equation (all masses here are rest masses): The calculation needs more time, more paper and more patience. We will only show you the result of the simplest case m1 = m2 = m:

  9. The difference of nonrelativistic and relativistic calculation is visible on the graph: Nonrelativistic case The quantitative understanding of kinematics of collisions enables us to compare masses of coins or particles by just colliding them. First success! Collisions are good at least for something, not only for destroying everything ...

  10. The formula for relativistic energy can be rewritten in the form Energy and momentum have different values in different reference frames - a bottle in my hand in the train has no kinetic energy with reference to the train but it may have quite significant energy referred to the ground. But the special expression above made from E and p equals always (independently of the reference frame) the square of the particle rest mass times c4, i.e. is constant. This feature offers a surprisingly simple way to measure themass of an unstable particle: Measure energies and momenta of decay products and , then calculate Unstable particle with unknown mass You have got the mass M ! You will find words like energy-momentum fourvector, invariant mass etc. in advanced textbooks. They refer to the same things as above, you can learn more ...

  11. To measure the masses of particles is clearly not enough. How to get some deeper insight, how to understand structure, interactions etc.? Maybe the way is to look what everything can happen and how probably. We can start again with our macroworld. What is the probability that I will catch a ball shot at me? The best solution to answer this question is to make an experiment. Being exposed to randomly placed moderate shots I caught all the yellow shots. After several repetition of this experiment I succeeded catching everything inside the area with the yellow boundary.

  12. My “ability” to catch ball shots is characterized by the yellow area - it is about 2 m2 and it means that from the “flow” of shots with a density 10 shots per square meter I expect to have 20 catches. My “catching ability” is characterized by an effective area which I can cover. Physicists use a special name for this quantity - “the cross section” and typically use the letter s for it. If the flow of incoming particles with the density j hits the target, then the number of interesting events N with the cross section s is The standard unit for the cross section is 1 barn = 1 b = 10-28 m2

  13. In terms of cross-section we can express a lot of information. For example the shots I am catching can vary in hardness - in energy of the ball. I can easily catch the slow balls but I will probably try to hide myself from hard shots. So for hard shots my cross section of catching will be zero and the cross section of a ball hitting me will be close to the area of my silhouette (+ the band around accounting for the diameter of the ball). One may consider the special (or “partial”) cross section for a ball breaking my glasses etc. All the possible processes can be summarized in the total cross section.

  14. sball-me The energy dependence of different cross-section relating to the interaction of me and the ball could look like this graph (and betray a lot about me …) At this energy I try to avoid ball hit At this energy I am frozen from the fear total At this energy I am catching best catch hit injury death Energy of the ball in apropriate units The cross-sections for proton-proton interactions are displayed on this plot: Elastic crosssection - colliding particles stay intact, they only change the direction of their flight inelastic At this energy the colliding protons have enough energy to create a new particle - pion. This is one example of a process contributing to the inelastic cross section = stotal - selastic

  15. To be continued

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