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    EQ: K^– + p → K^+ + K^0 + X

    My solution:

    LHS strangeness = -1
    RHS strangeness= +2
    therefore, X must = -3 strangeness
    therefore, the quark structure must be (sss).

    Mark Scheme says:

    1*** (strangeness on LHS is -1)strangeness on RHS without X is +2 /strangeness of X is -3 (one mark)
    thus sss
    OR
    2*** strangeness on RHS without X is +2 / strangeness of X is -1(1 mark)
    thus sdd (2 marks)

    I don't understand the part labelled '2***' as strangeness would not be conserved! It can only be 1***, and why is solution 1*** only one mark?
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    I understand now, if you don't then lemme explain.

    EDIT: The question => 'If strangeness is conserved in the equation, determine, with explanation, the quark structure of X'.

    It's because K^0 has two potential quarks structures:
    [1] (anti-strange, down) or...
    [2] (anti-down, strange)

    therefore we get two possible strangeness values, where
    [1] = -1
    [2] = +1.

    Therefore, strangeness of X can be either -1 or -3, depending on which quark structure of K^0 you use. The question wants you to assume both structures are possible, which is not obvious if you haven't done anything like this before. Hope this makes sense and helps! =]
 
 
 
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