Question

A photon and an electron, each of \(10\,\text{eV}\) 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 :

• A. \(275\)
• B. \(\frac{2}{450}\)
• C. \(\frac{1}{250}\)
• D. \(225\)

Hint:

Photon momentum is \(E/c\). Electron momentum is \(\sqrt{2mE}\). Always convert eV into joule before calculation. Compare orders of magnitude carefully.

Answer 1

The Correct Option is A

Concept:

• Momentum of a photon is given by \(p=\frac{E}{c}\).

• Momentum of a non-relativistic electron is \(p=\sqrt{2mE}\).

• Both particles have the same energy of \(10\,\text{eV}\).

Step 1: Calculate photon momentum
\[ E=10\times 1.6\times10^{-19} =1.6\times10^{-18}\,\text{J} \] \[ P_{ph}=\frac{E}{c} =\frac{1.6\times10^{-18}}{3\times10^8} =5.33\times10^{-27}\,\text{kg m s}^{-1} \]

Step 2: Calculate electron momentum
\[ P_e=\sqrt{2mE} \] \[ =\sqrt{2\times9\times10^{-31}\times1.6\times10^{-18}} \] \[ =\sqrt{28.8\times10^{-49}} \] \[ =5.37\times10^{-24}\,\text{kg m s}^{-1} \]

Step 3: Find the ratio
\[ \frac{P_e}{P_{ph}} = \frac{5.37\times10^{-24}} {5.33\times10^{-27}} \approx 275 \]