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PhysicsModern PhysicsAtomic structureEasy2 minQB

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The minimum energy required by a hydrogen atom in ground state to emit radiation in Balmer series is nearly :

Answer:

Solution:

For Balmer series, transition is from $n = 3, 4, 5, 6, e t c$ to $n = 2$ .

For transition to $n = 2$ , the minimum state to transition from is $n = 3$ .

Therefore, energy required $Δ E = E_{3} - E_{1} = - \frac{13.6}{9} - (- 13.6) = 12.1 e V$

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:Atomic structure

⏱ 2mℹ️ Source: QB

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