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Simple Nature - Light and Matter

Simple Nature - Light and Matter

Simple Nature - Light and Matter

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so that he could attempt to do physics experiments.Chadwick’s experiment worked like this. Suppose you performtwo Rutherford-type alpha scattering measurements, first one witha gold foil as a target as in Rutherford’s original experiment, <strong>and</strong>then one with a copper foil. It is possible to get large angles ofdeflection in both cases, but as shown in figure l, the alpha particlemust be heading almost straight for the copper nucleus to get thesame angle of deflection that would have occurred with an alphathat was much farther off the mark; the gold nucleus’ charge is somuch greater than the copper’s that it exerts a strong force on thealpha particle even from far off. The situation is very much likethat of a blindfolded person playing darts. Just as it is impossibleto aim an alpha particle at an individual nucleus in the target, theblindfolded person cannot really aim the darts. Achieving a veryclose encounter with the copper atom would be akin to hitting aninner circle on the dartboard. It’s much more likely that one wouldhave the luck to hit the outer circle, which covers a greater numberof square inches. By analogy, if you measure the frequency withwhich alphas are scattered by copper at some particular angle, saybetween 19 <strong>and</strong> 20 degrees, <strong>and</strong> then perform the same measurementat the same angle with gold, you get a much higher percentage forgold than for copper.l / An alpha particle must beheaded for the ring on the frontof the imaginary cylindrical pipein order to produce scattering atan angle between 19 <strong>and</strong> 20 degrees.The area of this ringis called the “cross-section” forscattering at 19-20 ◦ because it isthe cross-sectional area of a cutthrough the pipe.In fact, the numerical ratio of the two nuclei’s charges can bederived from this same experimentally determined ratio. Using thest<strong>and</strong>ard notation Z for the atomic number (charge of the nucleusdivided by e), the following equation can be proved (example 6):Z 2 goldZ 2 copper=number of alphas scattered by gold at 19-20◦number of alphas scattered by copper at 19-20 ◦By making such measurements for targets constructed from all theelements, one can infer the ratios of all the atomic numbers, <strong>and</strong>since the atomic numbers of the light elements were already known,atomic numbers could be assigned to the entire periodic table. Ac-Section 8.2 The Nucleus 483

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