# Q2 and Q3 (Homework 5, Physics 226)

(a) A computation of either lifetime must produce a result of the form $t = C\hbar^a c^b \alpha^p m_e^q$, for some dimensionless coe cient C and exponents a, b, p and q. What are the values of a, b and q?

(b) Muonium is the name given to bound states of a muon and antimuon. Muons interact just like electrons, but are heavier than electrons by a factor of m =me 207. Predict the lifetime of ortho- and para-muonium (the lowest energy muonium bound states with S = 1 or S = 0, respectively).

De Broglie says

$λ=h/(|p ⃗|)$

Also we know that

$E=\sqrt{(p ⃗^2 c^2+(m_0 c^2 )^2 })$ so that $E=\sqrt{(h^2/λ^2 c^2+(m_0 c^2 )^2)}$

For electron $m_0 c^2=0.511 MeV$ so that

$E=\sqrt{((6.626*10^{-34})^2/(10^{-16})^2*(3*10^8)^2+(0.511*10^6*1.6*10^{-19})^2)}=$

Question 3

a)

$τ=C*ℏ^a*c^b*α^p*m^q$

$α$ is the fine structure constant (dimensionless),C is dimensionless also

Therefore

$seconds=(Joule*seconds)^a*(m/s)^b*kg^q$ and since $J=kg*m^2/s^2 (W=(mv^2)/2)$

$s=[kg^a*(m/s)^2a s^a]*(m/s)^b*kg^q$

And thus

$a=-q$ $2a=-b$ $1=a-2a-b$

Heisenberg says

$ΔE*Δt≥ℏ/2$ so that $τ~ℏ/2E~ℏ/m$

Therefore

$τ_μ/τ_e =m_e/m_μ$

So that

$τ_μ (S=0 or singlet)=m_e/m_μ *τ_e (S=0)=1/207*125 ps=0.604 ps$

$τ_μ (S=1 or triplet)=m_e/m_μ *τ_e (S=1)=1/207*139 ps=0.671 ps$