Answer
$\lambda_{max}=6.62\times10^{-16}m$
Work Step by Step
This photon has just enough energy to create the 2 masses, with zero kinetic energy. (It is a maximum wavelength because if the wavelength increases, the energy decreases and can no longer produce the particles.)
Use equation 27–6.
$$\lambda_{max}=\frac{hc}{E}=\frac{hc}{2mc^2}=\frac{h}{2mc}$$
$$\lambda_{max}=\frac{6.63\times10^{-34}J \cdot s}{2(1.67\times10^{-27}kg)(3.00\times10^8m/s)}$$
$$\lambda_{max}=6.62\times10^{-16}m$$
Incidentally, the photon initially had some momentum and the products do not, so this process must take place near other particles to conserve overall momentum.
