B

Step 1: Calculate the energy available for the electron's kinetic energy after photoemission.
The energy of the photon is given as E_{photon} = 5.4852 ext{ eV}.
The ionization potential of the Li atom is E_{ionization} = 5.3918 ext{ eV}.
The excess energy (which becomes kinetic energy) is calculated as follows:
K.E. = E_{photon} - E_{ionization} = 5.4852 ext{ eV} - 5.3918 ext{ eV} = 0.0934 ext{ eV}.

Step 2: Convert the kinetic energy to joules.
1 ext{ eV} = 1.6 imes 10^{-19} ext{ J}, thus: K.E. = 0.0934 ext{ eV} * 1.6 imes 10^{-19} ext{ J/eV} = 1.4944 imes 10^{-20} ext{ J}.

Step 3: Use the kinetic energy equation to find the speed of the emitted electron.
K.E. = \frac{1}{2} m_{e} v^{2}
Rearranging gives: v = \sqrt{\frac{2 K.E.}{m_{e}}} = \sqrt{\frac{2 (1.4944 \times 10^{-20})}{9.1 \times 10^{-31}}}
Calculating this gives v \approx 1.73 \times 10^{5} ext{ m/s}.

Step 4: Analyze momentum conservation for the movement of the Li ion.
Since momentum is conserved, we have: \( m_{e} v_{e} = m_{Li^{2+}} v_{Li^{2+}} \), where \( m_{Li^{2+}} \) = \( \frac{6.94}{3} \times 10^{-3} \text{ kg} \) (for Li^{2+})
Hence, v_{Li^{2+}} = \frac{m_{e} v_{e}}{m_{Li^{2+}}}.
Using the relation:
v_{Li^{2+}} = \frac{(9.1 \times 10^{-31})(1.73 \times 10^{5})}{(6.94/3) \times 10^{-3}} \approx 7.885 \times 10^{-5} ext{ m/s}.
Thus, the Li^{2+} moves at speed \approx 7.885 \times 10^{-5} ext{ m/s} in the opposite direction to the emitted electron.
Therefore, the answer is B.