Chemistry - Atomic Structure Question with Solution | TestHub

ChemistryAtomic StructureMiscellaneous/MixedEasy2 minPYQ_2020

ChemistryEasysingle choice

An $α$ -paticle approaches the target nucleus of copper (Z = 29) in such a way that the value of impact parameter is zero. The distance of closest approach will be

Options:

Answer:

Solution:

Distance of closest approach
$K \frac{q_{1} q_{2}}{r} = (K . E .)_{α}$
$K = \frac{1}{4 π ε_{0}}$
$\frac{1}{4 π ε_{0}} \frac{Z e \times 2 e}{r} = (K . E .)_{α}$
$\frac{1}{4 π ε_{0}} \frac{2 \times 29 e^{2}}{r} = (K . E .)_{α}$
$r = \frac{1}{2 π ε_{0}} \times \frac{29 e^{2}}{(K . E .)_{α}}$

Stream:NTA_ABHYASSubject:ChemistryTopic:Atomic StructureSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2020

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