C

Step 1: Calculate the excess energy of the photon after ionizing the electron.
The energy of the photon (E) is 5.4852 eV, and the ionization potential (I) is 5.3918 eV. The excess energy (E_{excess}) that is given to the electron after overcoming the ionization potential is:
$$ E_{excess} = E - I = 5.4852 ext{ eV} - 5.3918 ext{ eV} = 0.0934 ext{ eV} $$
Step 2: Convert the excess energy into joules (since 1 eV = 1.6 \times 10^{-19} J):
$$ E_{excess} = 0.0934 ext{ eV} \times 1.6 \times 10^{-19} ext{ J/eV} = 1.4944 \times 10^{-20} ext{ J} $$
Step 3: Calculate the speed of the emitted electron using the kinetic energy formula, which is equal to the excess energy:
$$ KE = \frac{1}{2} m_{e} v^{2} \implies v = \sqrt{\frac{2\times KE}{m_{e}}} $$
where \( m_{e} = 9.1 \times 10^{-31} \text{ kg} \). Thus:
$$ v = \sqrt{\frac{2 \times 1.4944 \times 10^{-20}}{9.1 \times 10^{-31}}} $$
Step 4: Calculate v:
$$ v = \sqrt{\frac{2.9888 \times 10^{-20}}{9.1 \times 10^{-31}}} = \sqrt{3.280 \times 10^{10}} \approx 5.73 \times 10^{5} \text{ m/s} $$
Step 5: By conservation of momentum, the momentum of the emitted electron must equal the momentum of the Li ion (Li+). Since the electron moves at right angles to the photon's direction:
- Momentum of the electron: \( p_{e} = m_{e} v \)
- Total momentum of the ion can be denoted as \( p_{Li} = M_{Li} V_{Li} \), where \( M_{Li} = 6.94 \text{ g} = 6.94 \times 10^{-3} \text{ kg} \)
Step 6: Setting the momentum equations based on conservation of momentum:
In the vertical direction, the momentum of the electron will equal the momentum of the ion:
$$ p_{electron} = p_{Li} \implies m_{e} v = M_{Li} V_{Li} $$
Therefore,
$$ V_{Li} = \frac{m_{e} v}{M_{Li}} = \frac{(9.1 \times 10^{-31})(5.73 \times 10^{5})}{6.94 \times 10^{-3}} $$
Step 7: Calculate \( V_{Li} \):
$$ V_{Li} \approx \frac{5.20 \times 10^{-25}}{6.94 \times 10^{-3}} \approx 7.49 \times 10^{-23} \text{ m/s} $$
Step 8: Therefore, the speed of the Li ion is very small due to its much larger mass compared to that of the electron. The direction of the Li ion will be in the opposite direction to that of the emitted electron.
Hence, the relevant answer reflects the speed and direction based on this analysis.