Answer
$(a)$ $ E_K = 3.044 \times 10^{-13} J = 1.902 MeV$
$(b)$ $ v= 1.35 \times 10^7 m/s$
Work Step by Step
$(a)$ The total energy of photon is calculated as follows
$ E = \frac{hc}{\lambda} $
$ E = \frac{(6.626 \times10^{-34} J.s)(2.99 \times 10^8 m/s)}{3.00 \times 10^{-13} m}$
$E = 6.604 \times 10^{-13}J $
The energy equation of photon goes by $ E = E_K +E_B$. And the binding energy of a deuteron is $2.224 MeV$
$E_B = 2.224 MeV \times 1.6 \times 10^{-19} J/eV$
$E_B =3.56 \times 10^{-13} J$
The kinetic energy is $ E_K = E - E_B$.
$ E_K = 6.604 \times 10^{-13}J - 3.56 \times 10^{-13} J $
$ E_K = 3.044 \times 10^{-13} J $
$ E_K = 1.902 MeV$
$(b)$ The particles shared the energy, taking kinetic energy from (a) and solve for the velocity, gives
$E_K/2 = \frac{1}{2} mv^2$
$v= \sqrt{\frac{2E_K/2}{m}}$
$v= \sqrt{\frac{3.044 \times 10^{-13} J }{1.6605 \times 10^{-27} kg}}$
$v= 1.35 \times 10^7 m/s$
