Question

What should be the order of arrangement of de-Broglie wavelength of electron ($\lambda_e$), an $\alpha$-particle ($\lambda_\alpha$) and proton ($\lambda_p$) given that all have the same kinetic energy?

• A. $\lambda_e<\lambda_p<\lambda_\alpha$
• B. $\lambda_e = \lambda_p = \lambda_\alpha$
• C. $\lambda_e>\lambda_p>\lambda_\alpha$
• D. $\lambda_e = \lambda_p>\lambda_\alpha$

Hint:

For same kinetic energy, the lighter the particle, the longer its de-Broglie wavelength.
Electron is the lightest, so it has the largest wavelength among the given options.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The de-Broglie wavelength (\(\lambda\)) of a particle with mass \(m\) and kinetic energy \(K\) is given by the relation between wavelength and momentum.
Step 2: Key Formula or Approach:
\[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} \]
Since all particles have the same kinetic energy (\(K\)) and \(h\) is constant, \(\lambda \propto \frac{1}{\sqrt{m}}\).
Step 3: Detailed Explanation:
Compare the masses of the three particles:
1. Electron mass (\(m_e\)) is approximately \(9.1 \times 10^{-31} \text{ kg}\).
2. Proton mass (\(m_p\)) is approximately \(1836 \times m_e\).
3. \(\alpha\)-particle mass (\(m_\alpha\)) is approximately \(4 \times m_p\).

Order of masses: \(m_e<m_p<m_\alpha\).
Since the wavelength is inversely proportional to the square root of the mass:
\[ \frac{1}{\sqrt{m_e}}>\frac{1}{\sqrt{m_p}}>\frac{1}{\sqrt{m_\alpha}} \]
Therefore, \(\lambda_e>\lambda_p>\lambda_\alpha\).
Step 4: Final Answer:
The correct order of de-Broglie wavelengths is \(\lambda_e>\lambda_p>\lambda_\alpha\).