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

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

' λ₁ ' is the wavelength of series limit of Lyman series, ' λ₂ ' is… | MHT CET 2021 (21 Sep, Shift 1)

Q43MHT CET 2021 · 21 SEP · SHIFT 1Medium

Hydrogen Spectrum & Spectral Series

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

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Answer: (a)

Lyman limit: $\frac{1}{\lambda_1} = R$ ; Lyman first line ( $2 \to 1$ ): $\frac{1}{\lambda_2} = \frac{3R}{4}$
Balmer limit: $\frac{1}{\lambda_3} = \frac{R}{4}$
$\frac{1}{\lambda_1}-\frac{1}{\lambda_2} = \frac{R}{4} = \frac{1}{\lambda_3}$
So the answer is (a) $\frac{1}{\lambda_1}-\frac{1}{\lambda_2}=\frac{1}{\lambda_3}$ .

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Hydrogen Spectrum & Spectral Series question from MHT CET 2021 (21 Sep, Shift 1)

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