Decay width of the tau lepton

Question

The total decay width of a particle, $\Gamma$, is proportional to the fifth power of its mass, $m$, $$ \Gamma \propto m^5$$

We also know that the dominant decays of the tau lepton are $$\tau^- \to \nu_\tau e^- \nu_e$$ $$\tau^- \to \nu_\tau \mu^- \nu_\mu$$ $$\tau^- \to \nu_\tau \ du$$ with branching ratios of approximately 20%, 20% and 60%, respectively.

Given that the lifetime of a muon is known to be $\tau_\mu$ I have been told that the lifetime of the tau is

$$\tau_{\tau}=\frac{1}{5}\left(\frac{m_{\mu}}{m_{\tau}}\right)^{5} \tau_{\mu}=1.48 \times 10^{-7} \tau_{\mu}$$

I am unsure where the 1/5 factor has come from, any explanation would be appreciated.

Answer 1

The decay width $\Gamma=K m^5 $is an approximation that holds when all final state particles are much lighter than the decaying particle. For the muon, there is only one decay mode(the muon is not massive enough to create the lightest meson). So $\Gamma_\mu\approx K m_\mu^5 $. The tau has two leptonic decay modes, and one hadronic. The u and d quarks are colored and brings a factor of 3 to the decay rate. Thus $\Gamma_\tau \approx 5\times K m_\tau^5 $