JEE CHEMISTRY MCQ: STRUCTURE OF ATOM

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The frequency of radiation absorbed or emitted when transition occurs between two stationary states with energies

E_1

(lower) and

E_2

(higher) is given by

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Correct option: D

v=\frac{E_2-E_1}{h}

Option D (

v=\frac{E_2-E_1}{h}

) is correct because the frequency of emitted/absorbed radiation is proportional to the energy difference between states, divided by Planck's constant.

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