The energy of a photon is given by $E = pc$. Therefore, the momentum of the photon is: $$p_{Ph} = \frac{E}{c}$$
The kinetic energy of a non-relativistic electron is $E = \frac{p_e^2}{2m}$. Rearranging for momentum: $$p_e = \sqrt{2mE}$$
The ratio $\frac{p_e}{p_{Ph}}$ is: $$\frac{p_e}{p_{Ph}} = \frac{\sqrt{2mE}}{E/c} = \frac{c\sqrt{2mE}}{E} = c\sqrt{\frac{2m}{E}}$$
Convert $E = 20 \text{ eV}$ to Joules: $E = 20 \times 1.6 \times 10^{-19} \text{ J} = 3.2 \times 10^{-18} \text{ J}$. Substitute $c = 3 \times 10^8$, $m = 9 \times 10^{-31}$, and $E = 3.2 \times 10^{-18}$: $$\frac{p_e}{p_{Ph}} = 3 \times 10^8 \times \sqrt{\frac{2 \times 9 \times 10^{-31}}{3.2 \times 10^{-18}}}$$ $$\frac{p_e}{p_{Ph}} = 3 \times 10^8 \times \sqrt{\frac{18 \times 10^{-31}}{3.2 \times 10^{-18}}} = 3 \times 10^8 \times \sqrt{5.625 \times 10^{-13}} \approx 3 \times 10^8 \times 7.5 \times 10^{-7} \approx 225$$
Final Answer: 225
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