A beam of unstable neutral pions is observed to travel on average 8 m before they decay.If their proper lifetime is 0.8 $\times$ 10^-8 s,then calculate how fast the beam is traveling. A)0.96c B)0.92c C)0.29c D)0.98c E)3.5c

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Quizplus · Accepted Answer

The decay length $L$ in the lab frame is related to the proper lifetime $\tau_0$, the speed of the particle $v$, and the speed of light $c$ by the formula $L = v\gamma\tau_0$, where $\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$ is the Lorentz factor. Given $L = 8$ m and $\tau_0 = 0.8 \times 10^{-8}$ s, we can solve for $v$ by rearranging the formula to $v = \frac{L}{\gamma\tau_0}$. Since $L = v\tau$, where $\tau = \gamma\tau_0$ is the dilated lifetime, we find that $v = \frac{L}{\tau_0\gamma}$. Knowing that the pions travel 8 m before decaying and using the proper lifetime, we can infer that the speed must be such that the relativistic effects significantly increase the observed lifetime, allowing the pions to travel this distance. The correct speed that matches these conditions, given the options, is $0.96c$, which implies a significant relativistic time dilation allowing the pions to travel 8 m.

A Beam of Unstable Neutral Pions Is Observed to Travel ×\times×

A Beam of Unstable Neutral Pions Is Observed to Travel $\times$