ch44_p22 - antiquark since B = 1 then the quark composition...

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22. (a) Using Table 44-3, we find q = 0 and S = –1 for this particle (also, B = 1, since that is true for all particles in that table). From Table 44-5, we see it must therefore contain a strange quark (which has charge –1/3), so the other two quarks must have charges to add to zero. Assuming the others are among the lighter quarks (none of them being an
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Unformatted text preview: antiquark, since B = 1), then the quark composition is d s u . (b) The reasoning is very similar to that of part (a). The main difference is that this particle must have two strange quarks. Its quark combination turns out to be uss ....
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