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NEET 2015 (AIPMT) · Hydrogen Spectrum & Spectral Series

Hydrogen Spectrum & Spectral Series question from NEET 2015 (AIPMT)

Q56NEET 2015 · AIPMTMedium

Hydrogen Spectrum & Spectral Series

In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is

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Answer: (b)

Lyman longest ( $2 \to 1$ ): $\frac{1}{\lambda_L} = \frac{3R}{4}$
Balmer longest ( $3 \to 2$ ): $\frac{1}{\lambda_B} = \frac{5R}{36}$
$\frac{\lambda_L}{\lambda_B} = \frac{5/36}{3/4} = \frac{5}{27}$
So the answer is (b) $\frac{5}{27}$ .

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