A photon and an electron, each of 20 text{ eV | vedclass

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(C) For a photon, the energy $E$ and momentum $p_{ph}$ are related by $E = p_{ph}c$, so $p_{ph} = \frac{E}{c}$.For a non-relativistic electron, the ki

Vedclass · Accepted Answer

$225$

Vedclass · Answer

(C) For a photon, the energy $E$ and momentum $p_{ph}$ are related by $E = p_{ph}c$, so $p_{ph} = \frac{E}{c}$.For a non-relativistic electron, the kinetic energy $E$ and momentum $p_e$ are related by $E = \frac{p_e^2}{2m}$, so $p_e = \sqrt{2mE}$.The ratio of the momentum of the electron to that of the photon is $\frac{p_e}{p_{ph}} = \frac{\sqrt{2mE}}{E/c} = \frac{c\sqrt{2mE}}{E} = c\sqrt{\frac{2m}{E}}$.Given $E = 20 	ext{ eV} = 20 	imes 1.6 	imes 10^{-19} 	ext{ J} = 3.2 	imes 10^{-18} 	ext{ J}$.Substituting the values: $c = 3 	imes 10^8 	ext{ ms}^{-1}$, $m = 9 	imes 10^{-31} 	ext{ kg}$, $E = 3.2 	imes 10^{-18} 	ext{ J}$.$\frac{p_e}{p_{ph}} = 3 	imes 10^8 	imes \sqrt{\frac{2 	imes 9 	imes 10^{-31}}{3.2 	imes 10^{-18}}} = 3 	imes 10^8 	imes \sqrt{\frac{18 	imes 10^{-31}}{3.2 	imes 10^{-18}}} = 3 	imes 10^8 	imes \sqrt{5.625 	imes 10^{-13}} = 3 	imes 10^8 	imes \sqrt{56.25 	imes 10^{-14}} = 3 	imes 10^8 	imes 7.5 	imes 10^{-7} = 22.5 	imes 10^1 = 225$.

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