The $oldsymbol{ ext{Σ}^0}$ baryon, composed of the quark combination $uds$, is produced through the strong interaction between a $oldsymbol{ ext{π}^+}$ meson and a neutron - AQA - A-Level Physics - Question 11 - 2018 - Paper 1

To determine the quark composition of X in the reaction, we can analyze the conservation of baryon number and charge.

1. The $ext{π}^+$ meson has a charge of +1 and is made up of the quark combination uar{d}.
2. The neutron has a baryon number of +1 and is composed of the quark combination $udd$.
3. The $ext{Σ}^0$ baryon has a quark composition of $uds$ and a charge of 0.

Now let's apply conservation laws:

• The total charge before the reaction is $+1$ (from $ext{π}^+$) + $0$ (from neutron) = $+1$.
• After the reaction, $ext{Σ}^0$ also contributes 0 to the charge.
• Therefore, for the total charge to still equal +1, the charge of X must be +1.

Now, let's look at the baryon numbers:

• The baryon number before is +1 (neutron).
• The baryon number after including $ext{Σ}^0$ is also +1; thus, X must have a baryon number of 0.

Through these conservation principles, we find that:

• The quark composition of X must maintain neutral charge and a baryon number of 0, which indicates that it cannot contain any baryons. The only suitable option left is the quark combination uar{s} (option D), leading us to the conclusion that X is effectively u{ar{s}}.