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:
Matter waves (de Broglie waves) relate the momentum of a particle to its wavelength. We must express momentum in terms of kinetic energy and mass to compare the two particles.
Step 2: Key Formula or Approach:
De Broglie wavelength: \(\lambda = \frac{h}{p}\).
Kinetic energy: \(K = \frac{p^2}{2m} \implies p = \sqrt{2mK}\).
Substitute \(p\): \(\lambda = \frac{h}{\sqrt{2mK}}\).
Step 3: Detailed Explanation:
Since both particles have the same kinetic energy \(K\) and \(h\) is constant, the wavelength is inversely proportional to the square root of the mass:
\[ \lambda \propto \frac{1}{\sqrt{m}} \] We know that the mass of a proton (\(m_p\)) is much greater than the mass of an electron (\(m_e\)). (\(m_p \approx 1836 m_e\)).
Since \(m_p>m_e\), it follows directly from the inverse relationship that:
\[ \lambda_p<\lambda_e \] Or rewritten:
\[ \lambda_e>\lambda_p \] Step 4: Final Answer:
The correct relation is \(\lambda_e>\lambda_p\).