Given, uncertainty in momentum of an electron = 3 × uncertainty in position. What would be uncertainty in velocity of that electron.

3Δ x = Δ P ⇒ 3 ( Δ P ) 2 = 4 π h ⇒ ( Δ v ) 2 = 4 π ⋅ m 2 3 h ⇒ ( Δ v ) = 2 m 1 π 3 h

Chemistry - Atomic Structure Question with Solution | TestHub

ChemistryAtomic StructureHeisensberg's Uncertainty PrincipleMedium2 minQB

ChemistryMediumsingle choice

Given, uncertainty in momentum of an electron $= 3 \times$ uncertainty in position. What would be uncertainty in velocity of that electron.

Answer:

Solution:

$3Δ x = Δ P \Rightarrow \frac{( Δ P ) ^{2}}{3} = \frac{h}{4 π}$

$\Rightarrow (Δ v)^{2} = \frac{3 h}{4 π \cdot m ^{2}} \Rightarrow (Δ v) = \frac{1}{2 m} \frac{3 h}{π}$

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

⏱ 2mℹ️ Source: QB

Loading discussions...