Answer
$E_{min}=212MeV$
$\lambda=5.87\times10^{-15}m$
Work Step by Step
This photon has just enough energy to create the 2 muons, with zero kinetic energy.
$$E_{min}=2mc^2=2(207)(0.511MeV)=212MeV$$
Use equation 27–6.
$$\lambda=\frac{hc}{E}$$
$$\lambda=\frac{(6.63\times10^{-34}J \cdot s) (3.00\times10^8m/s)}{(211.6\times10^{6}eV)(1.60\times10^{-19}J/eV) }$$
$$\lambda=5.87\times10^{-15}m$$
