Question
A photon and an electron, each of 20 eV energy, move in free space. The ratio of linear momentum of electron $p_{e}$ to that of photon $p_{Ph},\frac{p_{e}}{p_{Ph}}$ is:
[Take speed of light $= 3 \times 10^{8}{\text{ }ms}^{- 1}$, charge of electron $= - 1.6 \times 10^{- 19}C$ and mass of electron $= 9 \times 10^{- 31}\text{ }kg$ ]
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Solution
Sol. [Key Concept: de Broglie wavelength — photon vs electron of equal energy]
$E_{e} = E_{Ph} = 20eV$
$$\begin{matrix} & \ \Rightarrow P_{e} = \sqrt{2m_{e}E_{e}},P_{ph} = \frac{E_{ph}}{c} \\ & \ \Rightarrow \frac{P_{e}}{P_{Ph}} = \frac{\sqrt{2m_{e}E_{e}}}{E_{Ph}}c = \sqrt{\frac{2m_{e}}{E_{Ph}}c} \\ & \frac{P_{e}}{P_{Ph}} = \sqrt{\frac{2 \times 9 \times 10^{- 31}}{20 \times 1.6 \times 10^{- 19}}} \times 3 \times 10^{8} \\ & \ = \frac{3}{4} \times 10^{- 6} \times 3 \times 10^{8} = \frac{900}{4} = 225 \end{matrix}$$
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