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NEET UG CHEMISTRY MCQ: STRUCTURE OF ATOM

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The wavelength of radiation emitted when electron falls from 4th Bohr's orbit to 2nd in H-atom is: (RH=−1.09678×10−7m−1)\left( {{R_\mathrm{H}} = - 1.09678 \times {{10}^{ - 7}}{\mathrm{m}^{ - 1}}} \right)
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Correct option: B
486 nm
Option B (486 nm) is correct because it is calculated using the Rydberg formula for the wavelength of emitted radiation.

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