In Lyman series, series limit of wavelength is λ₁. The wavelength of first line of Lyman series is λ₂ and in Balmer series, the series limit of wavelength is λ₃. Then the relation between λ₁, λ₂ and λ₃ is (a) λ₁ = λ₂ + λ₃ (b) λ₂ = λ₁ + λ₃ (c) (1)/(λ₁) = (1)/(λ₂) - (1)/(λ₃) (d) (1)/(λ₁) - (1)/(λ₂) = (1)/(λ₃)

Answer: (d). (1)/(λ) = R((1)/(n₁²)-(1)/(n₂²)) Lyman limit ( ∞ → 1): (1)/(λ₁) = R; Lyman first line ( 2 → 1): (1)/(λ₂) = 3R/4 Balmer limit ( ∞ → 2): (1)/(λ₃) = R/4 (1)/(λ₁)-(1)/(λ₂) = R-3R/4 = R/4 = (1)/(λ₃) So the answer is (d) (1)/(λ₁)-(1)/(λ₂)=(1)/(λ₃).

In Lyman series, series limit of wavelength is λ₁. The wavelength of… | MHT CET 2023 (12 May, Shift 1)

Q32MHT CET 2023 · 12 MAY · SHIFT 1Medium

Hydrogen Spectrum & Spectral Series

In Lyman series, series limit of wavelength is $\lambda_1$ . The wavelength of first line of Lyman series is $\lambda_2$ and in Balmer series, the series limit of wavelength is $\lambda_3$ . Then the relation between $\lambda_1, \lambda_2$ and $\lambda_3$ is

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Answer: (d)

$\frac{1}{\lambda} = R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
Lyman limit ( $\infty \to 1$ ): $\frac{1}{\lambda_1} = R$ ; Lyman first line ( $2 \to 1$ ): $\frac{1}{\lambda_2} = \frac{3R}{4}$
Balmer limit ( $\infty \to 2$ ): $\frac{1}{\lambda_3} = \frac{R}{4}$
$\frac{1}{\lambda_1}-\frac{1}{\lambda_2} = R-\frac{3R}{4} = \frac{R}{4} = \frac{1}{\lambda_3}$
So the answer is (d) $\frac{1}{\lambda_1}-\frac{1}{\lambda_2}=\frac{1}{\lambda_3}$ .

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Hydrogen Spectrum & Spectral Series question from MHT CET 2023 (12 May, Shift 1)

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