Physics - Modern Physics Question with Solution | TestHub

For a photon, the energy $E$ and linear momentum $p_{Ph}$ are related by:

$p_{Ph}=\frac{E}{c}$

For an electron, the kinetic energy $E$ and linear momentum $p_e$ are related by:

$p_e=\sqrt{2mE}$

The ratio of the linear momentum of the electron to that of the photon is:

$\frac{p_e}{p_{Ph}}=\frac{\sqrt{2mE}}{\frac{E}{c}}=c\sqrt{\frac{2m}{E}}$

Given:
$E=20\text{ eV}=20\times 1.6\times 10^{-19}\text{ J}=32\times 10^{-19}\text{ J}$
$m=9\times 10^{-31}\text{ kg}$
$c=3\times 10^8\text{ m/s}$

Substituting the values into the ratio:

$\frac{p_e}{p_{Ph}}=3\times 10^8\times \sqrt{\frac{2\times 9\times 10^{-31}}{32\times 10^{-19}}}$

$\frac{p_e}{p_{Ph}}=3\times 10^8\times \sqrt{\frac{18\times 10^{-31}}{32\times 10^{-19}}}$

$\frac{p_e}{p_{Ph}}=3\times 10^8\times \sqrt{\frac{9}{16}\times 10^{-12}}$

$\frac{p_e}{p_{Ph}}=3\times 10^8\times \left(\frac{3}{4}\times 10^{-6}\right)$

$\frac{p_e}{p_{Ph}}=\frac{9}{4}\times 10^2=2.25\times 100=225$