Answer
$p = 6.45\times 10^{-27}~kg~m/s$
Work Step by Step
In pat (a), we found that the energy of the photon that is emitted is $~~12.1~eV$
We can find the momentum of the photon:
$p = \frac{E}{c}$
$p = \frac{(12.1~eV)(1.6\times 10^{-19}~J/eV)}{3.0\times 10^8~m/s}$
$p = 6.45\times 10^{-27}~kg~m/s$
