chap45, p45 026

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26.

(a) Using Table 45-3, 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 45-5, 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 antiquark, since B = 1), then the quark
composition is ¯

u¯s¯

d.

(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 ¯

u¯s¯s.


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