Chemistry - Atomic Structure Question with Solution | TestHub

ChemistryAtomic StructureHeisensberg's Uncertainty PrincipleMedium2 minQB

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Passage / Comprehension

Both the de Broglie hypothesis and Heisenberg's uncertainty principle establish that microscopic particles have a more prominent wave nature.

Heisenberg realized that, in the world of very small particles, one cannot measure any property of a particle without interacting with it in some way. This introduces an unavoidable uncertainty into the result. One can never measure all properties exactly. The minimum uncertainty in the position of any body is given by the wavelength of light. Heisenberg’s uncertainty principle states: $Δ x Δ P \geq \frac{h}{4 π} = \frac{ℏ}{2}$

Starting from uncertainity in position and momentum in Heisenbergs uncertainity principle, what will be the correct relation of uncertainity in de-Broglie wavelength and position of a particle in one dimension?

Options:

Answer:

Solution:

$Δ x \cdot Δ p ∵ Δ p ∴ Δ p \cdot Δ x ∴ Δ λ \cdot Δ x \geq \frac{h}{4 π} \dots (i) λ = \frac{h}{p} \dots (ii) = (\frac{dp}{d λ}) Δ λ = - \frac{h}{λ ^{2}} \cdot Δ λ = - \frac{h}{λ ^{2}} .Δ λ \cdot Δ x \geq \frac{h}{4 π} \geq - \frac{λ ^{2}}{4 π}$

Stream:JEESubject:ChemistryTopic:Atomic StructureSubtopic:Heisensberg's Uncertainty Principle

⏱ 2mℹ️ Source: QB

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