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Chapter 39: Problem 24

Are either or both of these decay schemes possible for the tau particle: \((\mathrm{a}) \tau^{-} \rightarrow e^{-}+\bar{\nu}_{e}+\nu_{\tau} ;\) (b) \(\tau^{-} \rightarrow \pi^{-}+\pi^{0}+\nu_{\tau} ?\)

Short Answer

Expert verified
Yes, both decay schemes a) \(\tau^{-} \rightarrow e^{-}+\bar{\nu}_{e}+\nu_{\tau}\) and b) \(\tau^{-} \rightarrow \pi^{-}+\pi^{0}+\nu_{\tau}\) are possible. In both decays, lepton number, charge, and baryon numbers are conserved.

Step by step solution

01

Decay Scheme (a)

For decay scheme (a), the primary considerations are the conservation of lepton number and electric charge. To start, looking at the lepton numbers: The left-hand side (LHS) has a lepton number of 1 (due to the existence of a single tau lepton), whereas the right-hand side (RHS) also has a lepton number of 1 (consisting of \(e^{-}\) and \(\bar{\nu}_{e}\) combined). Hence, lepton numbers are conserved. Next, considering the electric charge: The charge on the LHS is -1 (from \(\tau^{-}\)) and on the RHS is also -1 (from \(e^{-}\)), indicating that the charge is also conserved. There are no hadrons present before or after decay, so baryon number is also conserved. Hence, the decay is possible.
02

Decay Scheme (b)

For decay scheme (b), the considerations are again lepton number, electric charge and baryon number. First, looking at lepton numbers:The LHS has lepton number 1 (from \(\tau^{-}\)), and on the RHS the number is also 1 (from \(\nu_{\tau}\)). So, lepton numbers are conserved.Next, considering the electric charge:On the LHS, the charge is -1 and on the RHS, there is a total charge of -1 (\(-1\) from \(\pi^{-}\) and \(0\) from \(\pi^{0}\)). Hence, charge is conserved.There are, again, no hadrons present before or after decay, and so baryon number is conserved. Detecting \( \pi^{0} \rightarrow \gamma + \gamma \) is possible experimentally. Therefore, the decay is also possible.

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