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JEE CHEMISTRY MCQ: STRUCTURE OF ATOM

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Calculate the wave number for the shortest wavelength transition in the Balmer series of atomic hydrogen.

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Correct option: A

27419.25 \mathrm{~cm}^{-1}

Option A (

27419.25 \mathrm{~cm}^{-1}

) is correct because the shortest wavelength corresponds to the largest energy transition, which in the Balmer series is from infinity to n=2.

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