Q12.

The equation shows an interaction between a proton and a negative kaon that results in the formation of particle, X.

K^– + p → K⁺ + K⁰ + X

(a)

(i) State and explain whether X is a charged particle.

Yes, it is charged - it is negative.

	K–	+	p	→	K+	+	K0	+	X
charge conservation	-1	+	+1	→	+1	+	0	+	-1

 

The above calculation of how charge must balance clearly shows the examiner you understand what you are doing!

The reactants are negative and positive - resulting in a zero charge - therefore to conserve charge the products must also result in zero overall charge.

[2 marks]

(ii) State and explain whether X is a lepton, baryon or meson.

X must be a baryon.

	K–	+	p	→	K+	+	K0	+	X
baryon conservation	0	+	+1	→	0	+	0	+	+1

 

[2 marks]

(iii) State the quark structure of the K^–, K⁺ and the K⁰.

kaon	quark	antiquark
K–	s	̅u̅
K+	u	̅s̅
K0	s	̅d̅
or K0	d	̅s̅
In each case the symbols or words are acceptable but there must be a bar over anti - quark if symbols are used

 

[3 marks]

(iv) Strangeness is conserved in the interaction. Determine, explaining your answer, the quark structure of X.

We have two possible quark combinations for K₀ - therefore we have two possible outcomes...

	K–	+	p	→	K+	+	K0	+	X	quark composition
strangeness conservation	-1	+	0	→	+1	+	+1	+	-3	sss
or	-1	+	0	→	+1	+	-1	+	-1	dds

 

quark transfer method 1:

su̅ + uud →us̅ + sd̅ + X

cancelling the quarks and same flavour antiquark on each side we get:

s + ud → u + d̅ + X

cancelling the quarks that undergo no change

s + d → d̅ + X

X must therefore be dds

quark transfer method 2:

su̅ + uud →us̅ + ds̅ + X

cancelling the quarks and same flavour antiquark on each side we get:

s + ud →us̅ + ds̅ + X

cancelling the quarks that undergo no change

s →s̅ + s̅ + X

X must therefore be sss

[3 marks]