Pair Production

$E = hv$

$v$ = $\frac{c}{\lambda}$

$\textit{Minimum energy}$

= $2m_{e}v^2$

= $2(9.1 * 10^{-31} kg)$ $(3 * 10^8 m/s)^2$

= $1.638 * 10^{-13} J$

= $1.022$ $MeV$

$\textit{Minimum frequency}$

= $\frac{2m_{e}v^2}{h}$

= $\frac{2(9.1 * 10^{-31} kg)(3 * 10^8 m/s)^2}{6.626 * 10^{-34} J s}$

= $2.472 * 10^{20}$ $Hz$

$\textit{Minimum wavelength}$

= $\frac{c}{frequency}$

= $\frac{3 * 10^8 m/s}{2.472 * 10^{20} Hz}$

= $1.2 * 10^{-12}$ $m$

This is the threshold wavelength of an incoming photon in pair production. Simply use the photoelectric effect formula of hv and set it equal to two electron (neglecting nuclear recoil)...and solve in terms of wavelength! :)