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PhysicsModern PhysicsAtomic structureEasy2 minQB

PhysicsEasysingle choice

The ratio of the magnitude of the kinetic energy to the potential energy of an electron in the $5^{th}$ excited state of a hydrogen atom is :

Answer:

Solution:

As we know, $K E = \frac{1}{2} ∣ PE ∣$ for any orbit.

$∴ \frac{K E}{∣ PE ∣} = \frac{1}{2}$

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:Atomic structure

⏱ 2mℹ️ Source: QB

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