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

Step 1: Note the common energy.
Both the photon and the electron carry the same energy, $E = 10\,\text{eV}$. Let us convert this into joules so every quantity is in SI units.
\[ E = 10 \times 1.6\times10^{-19} = 1.6\times10^{-18}\,\text{J} \]
Step 2: Momentum of the photon.
A photon is massless, so its momentum follows directly from its energy: $p_{ph} = E/c$.
\[ P_{ph} = \frac{1.6\times10^{-18}}{3\times10^{8}} = 5.33\times10^{-27}\,\text{kg m s}^{-1} \]
Step 3: Momentum of the electron.
At $10\,\text{eV}$ the electron is non-relativistic, so we use $E = p^2/2m$, which rearranges to $p = \sqrt{2mE}$.
\[ P_e = \sqrt{2 \times 9\times10^{-31} \times 1.6\times10^{-18}} \]
Step 4: Evaluate the electron momentum.
\[ P_e = \sqrt{28.8\times10^{-49}} = 5.37\times10^{-24}\,\text{kg m s}^{-1} \]
Step 5: Form the ratio.
Divide the electron momentum by the photon momentum.
\[ \frac{P_e}{P_{ph}} = \frac{5.37\times10^{-24}}{5.33\times10^{-27}} \]
Step 6: Final value.
Carrying out the division gives a number close to $275$. The electron, having mass, carries far more momentum than the photon at the same energy.
\[ \boxed{\dfrac{P_e}{P_{ph}} \approx 275} \]