# Q2 and Q3 (Homework 5, Physics 226)

2. (3 pts) To learn about nuclear structure from electron scattering, the de Broglie wavelength of the electron must be smaller than the size of the nucleus. How much energy must an electron have for its wavelength to be less than 0.1 fm?
3. (8 pts) The lifetime of ortho-positronium, the lowest energy spin-triplet (S = 1) bound state of an electron and a positron, is 139 ns. The lifetime of para-positronium, the lowest energy spin-singlet (S = 0) bound state of an electron and a positron, is 125 ps.

(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).

Question 2

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

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

$=1.9878*10^{-9} J=12.42 GeV$

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

$a=-q$  $2a=-b$  $1=a-2a-b$

or {
$a=1/3$
$b=2/3$
$q=-1/3$
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

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

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