$14.1.2.$ After $10^{-8}$ seconds following the passage of a $\pi^0$-meson through detector A, both detectors A and B registered gamma ($\gamma$) quanta, which were produced from the decay of the $\pi^0$-meson: $$\pi^0 \rightarrow \gamma + \gamma$$ The distance between the detectors is 1 meter. What was the speed of the $\pi^0$-meson?
Solution
Since the detectors simultaneously registered the gamma quanta, this means the quanta traveled equal distances (their speeds are equal and equal to $c$, the speed of light). Therefore, the decay point of the $\pi^0$-meson is in the middle between the detectors: $\frac{\ell}{2}$.
According to the problem, the total time from the meson's flight to the decay point and then the flight of the gamma quantum to the detector is $t = 10^{-8}$ s: $$\frac{\ell}{2} \left( \frac{1}{\upsilon} + \frac{1}{c} \right) = t$$ Here, $\upsilon$ is the unknown speed of the meson. Solving this equation for $\upsilon$: $$\upsilon = \frac{\ell c}{2ct - \ell}$$ Substituting the numerical values into the expression gives the speed of the $\pi^0$-meson: $$\upsilon = 6 \cdot 10^{7} \text{ m/s}$$
