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A difference of 2.3 eV separates two ene | Class 12 Physics Chapter Atoms, Atoms NCERT Solutions

Q4.

A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level?

Separation of two energy levels in an atom, E = 2.3 eV = 2.3 × 1.6 × 10 −19 = 3.68 × 10 −19 J

Let ν be the frequency of radiation emitted when the atom transits from the upper level to the lower level.

We have the relation for energy as: E = hv

Where, h = Planck’s constant = 6.62 x 10-34 Js

∴ V = E/h = 3.68 x 10-19 / 6.62 x 10-32 = 5.55 x 1014 Hz

Hence, the frequency of the radiation is 5.6 × 10 14 Hz.

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Important Questions & Answers

What is the correct answer to: A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level??

Separation of two energy levels in an atom, E = 2.3 eV = 2.3 × 1.6 × 10 −19 = 3.68 × 10 −19 J

Let ν be the frequency of radiation emitted when the atom transits from the upper level to...

How do you solve A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level? step by step?

Step-by-step explanation:
• Separation of two energy levels in an atom, E = 2
• 3 eV = 2
• 3 × 1
• 6 × 10 −19 = 3
• 68 × 10 −19 J

Why is this answer important for exams?

This question is important because it tests key concepts from the NCERT syllabus and is frequently asked in CBSE exams.

Which NCERT concept is used in this question?

This question is based on core NCERT concepts explained in the chapter and should be revised thoroughly before exams.

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Students often lose marks by skipping steps, writing incomplete explanations, or misunderstanding keywords used in the question.

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