Date | November 2015 | Marks available | 2 | Reference code | 15N.3.SL.TZ0.13 |
Level | Standard level | Paper | Paper 3 | Time zone | Time zone 0 |
Command term | Show that | Question number | 13 | Adapted from | N/A |
Question
This question is about interactions and quarks.
For the lambda baryon \({\Lambda ^0}\), a student proposes a possible decay of \({\Lambda ^0}\) as shown.
\[{\Lambda ^0} \to p + {K^ - }\]
The quark content of the \({K^ - }\) meson is \({\rm{\bar us}}\).
A lambda baryon \({\Lambda ^0}\) is composed of the three quarks uds. Show that the charge is 0 and the strangeness is \( - 1\).
Discuss, with reference to strangeness and baryon number, why this proposal is feasible.
Strangeness:
Baryon number:
Another interaction is
\[{\Lambda ^0} \to p + {\pi ^ - }\]
In this interaction strangeness is found not to be conserved. Deduce the nature of this interaction.
Markscheme
\( + \frac{2}{3} - \frac{1}{3} - \frac{1}{3} = 0\) for charge;
any particle containing a strange quark has strangeness of \( - 1\);
strangeness:
the \(p\) has a strangeness of 0;
the \({K^ - }\) particle has a strangeness of \( - 1\);
baryon number:
lambda and protons are baryons each having a baryon number of \( + 1\);
the \({K^ - }\) meson has a baryon number of 0;
only during the weak interaction strangeness is not conserved (therefore it is a weak interaction);
Examiners report
Well answered question, often very clearly and straightforward; some, even better candidates made mistakes in calculation in (b)(iii). SL candidates showed more difficulty with (b)(iii), often using an incorrect approach.
Well answered question, often very clearly and straightforward; some, even better candidates made mistakes in calculation in (b)(iii). SL candidates showed more difficulty with (b)(iii), often using an incorrect approach.
Well answered question, often very clearly and straightforward; some, even better candidates made mistakes in calculation in (b)(iii). SL candidates showed more difficulty with (b)(iii), often using an incorrect approach.