If the wavelength of the first member of Lyman series of hydrogen is λ. The wavelength of the second member will be (a) 27/5λ (b) 5/27λ (c) 27/32λ (d) 32/27λ

Answer: (c). First member ( 2 → 1): (1)/(λ) = 3R/4 Second member ( 3 → 1): (1)/(λ') = R(1-1/9) = 8R/9 (λ')/(λ) = (3/4)/(8/9) = 27/32 So the answer is (c) 27/32λ.

If the wavelength of the first member of Lyman series of hydrogen is… | JEE Main 2024 (31 Jan, Shift 1)

Q24JEE Main 2024 · 31 JAN · SHIFT 1Medium

Hydrogen Spectrum & Spectral Series

If the wavelength of the first member of Lyman series of hydrogen is $\lambda$ . The wavelength of the second member will be

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Answer: (c)

First member ( $2 \to 1$ ): $\frac{1}{\lambda} = \frac{3R}{4}$
Second member ( $3 \to 1$ ): $\frac{1}{\lambda'} = R\left(1-\frac{1}{9}\right) = \frac{8R}{9}$
$\frac{\lambda'}{\lambda} = \frac{3/4}{8/9} = \frac{27}{32}$
So the answer is (c) $\frac{27}{32}\lambda$ .

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