Question

Question: A kaon (denoted ) is an unstable particle of mass $\mathbf{8}\mathbf{.}\mathbf{87}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{28}}\mathbf{}\mathit{k}\mathit{g}$. One of the means by which it decays is by spontaneous creation of two pions, and . The decay process may be represented as

${\mathit{K}}^{\mathbf{0}}\mathbf{}\stackrel{}{\mathbf{\to }}\mathbf{}{\mathit{\pi }}^{\mathbf{+}}\mathbf{}\mathbf{+}\mathbf{}{\mathit{\pi }}^{\mathbf{-}}$

Both the ${\mathit{\pi }}^{\mathbf{+}}$ and ${\mathit{\pi }}^{\mathbf{-}}$ have mass $\mathbf{2}\mathbf{.}\mathbf{49}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{28}}\mathbf{}\mathbf{kg}$. Suppose that a kaon moving in the + x direction decays by this process, with the ${\mathit{\pi }}^{\mathbf{+}}$ moving off at speed 0.9C and the ${\mathit{\pi }}^{\mathbf{-}}$ at 0.8c.

(a) What was the speed of the kaon before decay?

(b) In what directions do the pions move after the decay?

Short answer

Answer:

1. The kaon particle is moving along the positive x-axis at a speed of .
2. It decays into two charged pions ${\pi }^{+}$and ${\pi }^{-}$, with ${\pi }^{+}$ traveling at 0.9c in the direction from40⁰ the horizontal and ${\pi }^{-}$travelling at 0.8c in the direction -84.6⁰ from the horizontal.

Step-by-step solution

Define kaon particle:

In particle physics, a kaon denoted by K is any of a group of four mesons distinguished by a quantum number called strangeness.

A neutral kaon is denoted by $K^0$. It decays into two charged pions ${\pi }^{+}$ and ${\pi }^{-}$. Now suppose a neutral kaon $K^0$ is moving along positive x-direction and decays into two pions as shown in the figure below. One pion ${\pi }^{+}$ moves off at 0.9c and the other pion ${\pi }^{-}$ at 0.8c .

(a) Apply Energy conservation to find the velocity of kaon before decay:

Apply the energy conservation as below.

$$\begin{gathered} {\gamma }_{K^0}{m}_{K^0}{c}^2={\gamma }_{0.9c}{m}_{\pi }{c}^2+{\gamma }_{0.8c}{m}_{\pi }{c}^2 \\ {\gamma }_{K^0}=\left(\frac{\left({\gamma }_{0.9c}+{\gamma }_{0.8c}\right)m_{\pi }}{m_{K^0}}\right) \end{gathered}$$

Finding the Lorentz factors $\left(\gamma =\frac{1}{\sqrt{1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}}\right)$for two pions and putting them in the above equation you get,$$\begin{gathered} {\gamma }_{K^0}=1.11 \\ {u}_{K^0}=0.4384c \end{gathered}$$

(b) Apply momentum conservation separately in x and y direction:

In x-direction,

$$\begin{gathered} {\gamma }_{K^0}{m}_{K^0}{u}_{K^0}={\gamma }_{0.9c}{m}_{\pi }\left(\left(0.9c\right)\cos {\theta }_1\right)+{\gamma }_{0.8c}{m}_{\pi }\left(\left(0.8c\right)\cos {\theta }_2\right) \\ 1.11\left(8.87\times {10}^{-28}kg\right)\left(0.4384c\right)=\left[2.29\times 0.9c\cos {\theta }_1+1.67\times 0.8c\cos {\theta }_2\right]\left(2.49\times {10}^{-28}kg\right) \\ 2.06\cos {\theta }_1+1.34\cos {\theta }_2=1.73 \end{gathered}$$

$1.73-2.06\cos {\theta }_1=1.34\cos {\theta }_2$ ..… (1)

In y-direction,

$$\begin{gathered} 0={\gamma }_{0.9c}{m}_{\pi }\left(0.9c\sin {\theta }_1\right)-{\gamma }_{0.8c}{m}_{\pi }\left(0.8c\sin {\theta }_2\right) \\ 2.06\sin {\theta }_1=1.34\sin {\theta }_2 \end{gathered}$$ ….. (2)

Squaring both the equations and adding them yields the values of angles as

${\theta }_1=40.36°$ and ${\theta }_2=84.56°$

Hence, ${\pi }^{+}$ is traveling at 0.9c in the direction 40⁰C from the horizontal and ${\pi }^{-}$ is travelling in the direction$-84.6°$ from the horizontal.