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Class 12
Physics
Atoms
a difference of 2 3 ev separates two energy levels...

Question:
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?

Answer:

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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Question:
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?

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Question: 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?

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Question:
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?