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: Formula
- Electron: $\lambda_e = \frac{h}{mv_e}$. $E_e = \frac{1}{2}mv_e^2$.
- Photon: $E_p = \frac{hc}{\lambda_p}$.

Step 2: Analysis
Given $\lambda_p = 2\lambda_e \implies \lambda_p = 2(\frac{h}{mv_e})$.
$E_p = \frac{hc}{2h/mv_e} = \frac{m v_e c}{2}$.

Step 3: Calculation
$\frac{E_p}{E_e} = \frac{m v_e c / 2}{m v_e^2 / 2} = \frac{c}{v_e}$.
Given $v_e = c/100 \implies \frac{E_p}{E_e} = \frac{c}{c/100} = 100$.

Step 4: Conclusion
Hence, the ratio is $10^2$. Final Answer: (B)