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Chapter 3: Problem 3

Calculate the branching ratio for the decay \(\mathrm{K}^{+} \rightarrow \pi^{+} \pi^{0}\), given the partial decay width \(\Gamma\left(\mathrm{K}^{+} \rightarrow \pi^{+} \pi^{0}\right)=\) \(1.2 \times 10^{-8} \mathrm{eV}\) and the mean \(\mathrm{ka}\) on lifetime \(\tau\left(\mathrm{K}^{+}\right)=1.2 \times 10^{-8} \mathrm{~s}\).

Short Answer

Expert verified
The branching ratio is 0.144

Step by step solution

01

- Understand the Formula

The branching ratio (BR) for a particular decay mode of a particle is given by the ratio of the partial decay width (\text{\( Γ ( K^+ →π ^ + π ^ 0 ) } )\) to the total decay width (\text{\(Γ_{ total } } \)). We can express it as:\text{\(BR=\frac{Γ(K^+ →π^ + π^ 0 )}{Γ_{ total }}\)}
02

- Calculate the Total Decay Width

The total decay width (\text{\(Γ_{ total } } )\) is connected to the mean lifetime (\text{\(τ(K^+ )}\)) of the particle. The relationship is given by:\text{\(Γ_{ total }=\frac{ 1 }{τ(K ^+ ) } \)}Given that:τ(K^+) = 1.2 × 10^−8 s,we can calculate Γtotal as follows:\text{\(Γ_{ total }=\frac{ 1 }{ 1.2×10^{-8} s } = 8.333× 10^{−8} eV}\)}
03

- Plug in Values and Calculate the Branching Ratio

Now, having found \text{\(Γ_{ total}\)} and given the partial decay width \text{\(Γ( K^+ →π^ + π^0 )\),we can substitute the values into the branching ratio formula:\text{\(BR=\frac{1.2× 10 ^{ −8 } eV}{ 8.33 × 10 ^{−8} eV }=0.144\)}

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

partial decay width
Partial decay width refers to the probability per unit time that a particle will decay in a particular way. It's essentially a measure of how 'likely' a specific decay mode is compared to others.
Each possible decay channel of a particle has a partial decay width and they add up to the total decay width of the particle. For the kaon decay example, the partial decay width \(Γ(K^+ → π^+ π^0) \) is given as \(1.2 × 10^{-8} eV \).
This tells us how fast the kaon decays into a charged pion and a neutral pion specifically, but not into other particles.
kaon lifetime
The kaon lifetime represents the average time a kaon particle exists before decaying. It is inversely proportional to the total decay width.
For example, the given lifetime \(τ(K^+) = 1.2 × 10^{-8} s\) means that, on average, a kaon exists for about that much time before it decays. This time is key to finding the total decay width of the particle and crucial in determining branching ratios.
branching ratio calculation
Branching ratio is a measure used in particle physics to express the fraction of particles decaying via a particular decay mode compared to the total decay.
It is calculated by dividing the partial decay width of a specific decay channel by the total decay width of the particle. Using the equation:
\[BR=\frac {Γ(K^+ →π^+ π^0)}{Γ_{total}} \]
You input the partial decay width \(1.2× 10 ^{ −8 } eV\) and the total decay width calculated as \(8.333× 10^{−8} eV\)
from the lifetime inverse. Plugging these values in, you get a branching ratio of \(0.144\).
particle physics
Particle physics is a field of study focused on the smallest known building blocks of the universe and the forces acting upon them. When studying decays, as in the exercise problem, physicists analyze how particles like kaons transform into other particles.
They use concepts like partial decay width, kaon lifetime, and branching ratios to understand the statistical nature and behaviors of these transformations. Understanding these concepts helps physicists predict the products of decays and the likelihood of their occurrences, forming the groundwork for much of modern particle physics theory and experimentation.

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Most popular questions from this chapter

At a future \(\mathrm{e}^{+} \mathrm{e}^{-}\)linear collider operating as a Higgs factory at a centre-of-mass energy of \(\sqrt{5}=250 \mathrm{GeV}\), the cross section for the process \(\mathrm{e}^{+} \mathrm{e}^{-} \rightarrow \mathrm{HZ}\) is \(250 \mathrm{fb}\). If the collider has an instantaneous luminosity of \(2 \times 10^{34} \mathrm{~cm}^{-2} \mathrm{~s}^{-1}\) and is operational for \(50 \%\) of the time, how many Higgs bosons will be produced in five years of running? Note: 1 femtobarn \(\equiv 10^{-15} \mathrm{~b}\).

For the process \(a+b \rightarrow 1+2\) the Lorentz-invariant flux term is $$ F=4\left[\left(p_{a} \cdot p_{b}\right)^{2}-m_{a}^{2} m_{b}^{2}\right]^{\frac{1}{2}} . $$ In the non-relativistic limit, \(\beta_{a} \ll 1\) and \(\beta_{b} \ll 1\), show that $$ F \approx 4 m_{a} m_{b}\left|\mathbf{v}_{a}-\mathbf{v}_{b}\right|, $$ where \(\mathbf{v}_{a}\) and \(\mathbf{v}_{b}\) are the (non-relativistic) velocities of the two particles.

The total \(\mathrm{e}^{+} \mathrm{e}^{-} \rightarrow \gamma \rightarrow \mu^{+} \mu^{-}\)annihilation cross section is \(\sigma=4 \pi \alpha^{2} / 3 s\), where \(\alpha \approx 1 / 137\). Calculate the cross section at \(\sqrt{s}=50 \mathrm{GeV}\), expressing your answer in both natural units and in barns \(\left(1\right.\) barn \(\left.=10^{-28} \mathrm{~m}^{2}\right)\). Compare this to the total pp cross section at \(\sqrt{5}=50 \mathrm{GeV}\) which is approximately \(40 \mathrm{mb}\) and comment on the result.

The Lorentz-invariant flux term for the process \(a+b \rightarrow 1+2\) in the centre-of-mass frame was shown to be \(F=4 p_{j}^{*} \sqrt{5}\), where \(p_{j}^{*}\) is the momentum of the intial-state particles. Show that the corresponding expression in the frame where \(b\) is at rest is $$ F=4 m_{b} p_{e} $$
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