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JEE Main 2020 (8 Jan, Shift 2) · Hydrogen Spectrum & Spectral Series

Hydrogen Spectrum & Spectral Series question from JEE Main 2020 (8 Jan, Shift 2)

Q47JEE Main 2020 · 8 JAN · SHIFT 2Medium

Hydrogen Spectrum & Spectral Series

The first member of the Balmer series of hydrogen atom has a wavelength of 6561 Å. The wavelength of the second member of the Balmer series (in nm) is:

Numerical answer type. Work it out, then open the answer.

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Answer: 486

$\frac{1}{\lambda_1} = \frac{5R}{36}$ ( $3 \to 2$ ), $\frac{1}{\lambda_2} = \frac{3R}{16}$ ( $4 \to 2$ )
$\lambda_2 = \frac{20}{27}\lambda_1 = \frac{20}{27}\times 6561 = 4860$ Å
Answer: 486 nm

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