Question

An electron and a proton possess same kinetic energy. If the de Broglie wavelengths of the electron and the proton are $\lambda_e$ and $\lambda_p$ respectively, identify the CORRECT relation.

• A. $\lambda_e > \lambda_p$
• B. $\lambda_e = \lambda_p$
• C. $\lambda_p = \sqrt{1836} \lambda_e$
• D. $\lambda_p = 1836 \lambda_e$
• E. $\lambda_p = 183 \lambda_e$

Hint:

At the same kinetic energy, the lighter particle (electron) always has the higher de Broglie wavelength.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
According to the de Broglie hypothesis, every moving particle has a wave associated with it, whose wavelength depends on its momentum.
Key Formula or Approach:
Wavelength in terms of kinetic energy ($K$): \[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} \]

Step 2: Detailed Explanation:
1. Both particles have the same kinetic energy $K$.
2. Thus, \( \lambda \propto \frac{1}{\sqrt{m}} \).
3. We know that the mass of a proton ($m_p$) is much greater than the mass of an electron ($m_e$):
\[ m_p \approx 1836 m_e \implies m_p > m_e \]
4. Since mass is in the denominator, the particle with the smaller mass will have the longer wavelength.
5. Therefore, \( \lambda_e > \lambda_p \).

Step 3: Final Answer:
The correct relation is $\lambda_e > \lambda_p$.