Question

An \( \alpha \)-particle, a proton, and an electron have the same kinetic energy. Which one of the following is correct in case of their De-Broglie wavelength:

• A. \( \lambda_{\alpha} < \lambda_{p} < \lambda_{e} \)
• B. \( \lambda_{\alpha} > \lambda_{p} > \lambda_{e} \)
• C. \( \lambda_{\alpha} = \lambda_{p} = \lambda_{e} \)
• D. \( \lambda_{\alpha} > \lambda_{p} < \lambda_{e} \)

Answer 1

The Correct Option is A

The De-Broglie wavelength is given by:

\[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} \]

For particles with the same kinetic energy:

\[ \lambda \propto \frac{1}{\sqrt{m}} \]

Since \( m_\alpha > m_p > m_e \), we have:

\[ \lambda_e > \lambda_p > \lambda_\alpha \]

Answer 2

The matter waves, the wavelength is associated with the microscopic particles like protons, electrons, neutrons, \(\alpha\)-particle etc., is or the order of \(10^{-10}m\).
The relation between de-Broglie wavelength \(\lambda\) and the kinetic energy \(K\) of the particle is given by:
\(λ = \frac{h}{m.v} = \frac{h}{√(2.m.K.E)}\)
\(\text{as K.E. is same } λ∝\frac{1}{\sqrt{m}}\)
mass of electron = \(9.1 × 10^{-31}\) kg
mass of proton = \(1.67 × 10^{-27}\) kg
mass of α-particle = \(6.68 × 10^{-27}\) kg
\(λ_e > λ_p > λ_α\)

So, the correct option is (A): \(λ_e > λ_p > λ_α\)