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JEE Main 2019 (9 Apr, Shift 1) · Hydrogen Spectrum & Spectral Series

Hydrogen Spectrum & Spectral Series question from JEE Main 2019 (9 Apr, Shift 1)

Q52JEE Main 2019 · 9 APR · SHIFT 1Medium

Hydrogen Spectrum & Spectral Series

Taking the wavelength of first Balmer line in hydrogen spectrum (n = 3 to n = 2) as 660 nm, the wavelength of the 2nd Balmer line (n = 4 to n = 2) will be:

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Answer: (b)

$\frac{1}{\lambda_{3\to2}} = \frac{5R}{36}$ , $\frac{1}{\lambda_{4\to2}} = \frac{3R}{16}$
$\lambda_{4\to2} = \frac{5/36}{3/16}\,\lambda_{3\to2} = \frac{20}{27}\times 660$
$\lambda_{4\to2} \approx 488.9$ nm
So the answer is (b) 488.9 nm.

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