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Chapter 12: Q31E (page 557)

Sketch Feynman vertex for the annihilation of a b quark and u¯.

Short Answer

Expert verified

The Feynman diagram is shown in the figure as follows:

Step by step solution

01

Given data

The u¯ has charge -23 and b has charge-13 .

02

Concept of Feynman diagram 

A Feynman diagram describes the interactions in quantum field theory. Solid lines and curved lines are used to represent different types of particle interactions.

03

Step 3:Sketch Feynman vertex

The Feynman diagram is shown in figure 1.


Figure 1

Because the type of quark changed, it is a weak interaction.

Because u¯ has charge -23 and b has charge-13 , they will exchange a W-boson.

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Most popular questions from this chapter

In the following exercises, two protons are smashed together in an attempt to convert kinetic energy into mass and new particles. Indicate whether the proposed reaction is possible. If not, indicate which rules are violated. Consider only those for charge, angular momentum, and baryon number If the reaction is possible, calculate the minimum kinetic energy required of the colliding protons.

p+p→p+p+p+p¯

(a) Determine the quark content of the antineutrons.

(b) Sketch the Feynman diagram for its β decay.

In non-relavistic quantum mechanics, governed by the Schrodinger equation, the probability of finding a particle does not change with time.

(a)

Prove it, Begin with the time derivative of the total probability

ddt∫Ψ*(x,t)Ψ'(x,t)dx=∫(Ψ(x,t)∂∂tΨ*(x,t)+Ψ*(x,t)∂∂tΨ(x,t))dx

Then use the Schrodinger equation to eliminate the partial time derivatives, integrate by parts, and show that the result is zero. Assume that the particle is well localised, so that ψ and ∂ψ∂xare 0 when evaluated at .±∞

(b) Does this procedure lead to the same conclusion if Wave function obeyKlein-Gordon rather than Shrodinger equation? Why and why not?

Sketch the Feynman diagram if the proposed decay is possible.

n¯→p+e-+v¯e

The electron mentioned in Section 12.3 for deep inelastic scattering experiments is20GeV , and the momentum is given as .20 GeV/c Why so simple a conversion?

See all solutions

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