Question

Let $E_{e}$ and $E_{p}$ represent kinetic energy of electron and photon. If de-Broglie wavelength of photon is twice that of electron, find $E_p/E_e$ (speed of electron $= c/100$)}

• A. 10
• B. $10^{2}$
• C. $10^{3}$
• D. $10^{4}$

Hint:

For an electron, $E_e = \frac{p^2}{2m}$. For a photon, $E_p = pc$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We compare the energy of a massive particle (electron) and a massless particle (photon) using their de-Broglie wavelengths.

Step 2: Key Formula or Approach:
1. Photon: \( E_p = \frac{hc}{\lambda_p} \).
2. Electron: \( E_e = \frac{1}{2}mv_e^2 \) and \( \lambda_e = \frac{h}{mv_e} \).

Step 3: Detailed Explanation:
Substitute \( m = \frac{h}{\lambda_e v_e} \) from the de-Broglie equation into the electron energy formula:
\[ E_e = \frac{1}{2} \left( \frac{h}{\lambda_e v_e} \right) v_e^2 = \frac{hv_e}{2\lambda_e} \]
Now, find the ratio:
\[ \frac{E_p}{E_e} = \frac{hc/\lambda_p}{hv_e/2\lambda_e} = \frac{2c}{v_e} \times \frac{\lambda_e}{\lambda_p} \]
Given \( \lambda_p = 2\lambda_e \implies \frac{\lambda_e}{\lambda_p} = \frac{1}{2} \).
Given \( v_e = \frac{c}{100} \implies \frac{c}{v_e} = 100 \).
\[ \frac{E_p}{E_e} = 2(100) \times \frac{1}{2} = 100 = 10^2 \]

Step 4: Final Answer:
The ratio of energies is \( 10^2 \).