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:
\( \lambda_{\alpha} < \lambda_{p} < \lambda_{e} \)
\( \lambda_{\alpha} > \lambda_{p} > \lambda_{e} \)
\( \lambda_{\alpha} = \lambda_{p} = \lambda_{e} \)
\( \lambda_{\alpha} > \lambda_{p} < \lambda_{e} \)
The Correct Option is A
Approach Solution - 1
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:
\[ \lambda_e > \lambda_p > \lambda_\alpha \]
Approach Solution -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): \(λ_e > λ_p > λ_α\)
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Dual Nature of Matter
- The concept of Dual Nature of Matter was proposed after various experiments supported both wave as well particle nature of light.
- The particle nature of matter came into the picture when Albert Einstein looked up to the experiment conducted by Max Planck and observed that the wavelength and intensity of matter have a certain impact on the ejected electrons. Experiments such as the photoelectric effect suggested that light has a particle nature, i.e. light travels in form of packets of energy (E = h\(\nu\))
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