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PhysicsModern PhysicsAtomic structureEasy2 minPYQ_2020
PhysicsEasysingle choice

Anα-particle after passing through a potential difference ofVvolts collides with a nucleus. If the atomic number of the nucleus isZthen the distance of closest approach ofα-particle to the nucleus will be

Options:

Answer:
A
Solution:

K.E. = P.E. = qV

⇒2eV=KZe2ed=qV

∴d=9×109×Z×e×2e2eV

∴d=9×109×1.6×10–19×ZV

d=14.4×10-10ZVm

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Modern PhysicsSubtopic:Atomic structure
⏱ 2mℹ️ Source: PYQ_2020

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