Chapter 32, Problem 60GP

(a)

To determine

To Identify: The missing particle in the given reaction.

Answer to Problem 60GP

The missing particle is ${\pi }^0=u\overline{u}+d\overline{d}$ .

Explanation of Solution

The given reaction is,

  $p+p\to p+n+{\pi }^{+}+?$

In this reaction the missing particle must be neutral according to the charge conservation. In the case of conservation of baryon number, the missing particle must be meson. But for the strangeness conservation, the missing particle must be made up of up and down quarks and antiquarks only. On the basis of above information, the missing particle is ${\pi }^0=u\overline{u}+d\overline{d}$

Conclusion:

Thus, the missing particle is ${\pi }^0=u\overline{u}+d\overline{d}$ .

(b)

To determine

To Explain: The missing particle in the given reaction.

Answer to Problem 60GP

The missing particle is $v_{\mu }$ .

Explanation of Solution

The given reaction is,

  $p+?\to n+{\mu }^{+}$

In the given reaction, one lepton is present. So, there is a weak interaction.

For the conservation, the missing particle must be neutral. For the conservation of muon lepton number, the missing particle must be an antiparticle in muon family.

Conclusion:

So, the missing particle is $v_{\mu }$ .