α-particle, proton & electron have same kinetic energy. Select correct order of their de-Broglie wavelength.
\(λ_e > λ_p > λ_α\)
\(λ_α > λ_e > λ_p\)
\(λ_p = λ_α = λ_e\)
\(λ_p > λ_e > λ_α\)
The Correct Option is A
Solution and Explanation
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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