In hydrogen emission spectrum, for any series, the principal quantum number is n. Corresponding maximum wavelength λ is ( R = Rydberg's constant) (a) (R(2n+1))/(n²(n+1)) (b) (n²(n+1)²)/(R(2n+1)) (c) (n²(n+1))/(R(2n+1)) (d) (R(2n+1))/(n²(n+1)²)

Answer: (b). Maximum wavelength of a series ending on level n comes from (n+1) → n. (1)/(λ) = R(1/n²-(1)/((n+1)²)) = (R(2n+1))/(n²(n+1)²) λ = (n²(n+1)²)/(R(2n+1)) So the answer is (b).

In hydrogen emission spectrum, for any series, the principal quantum… | MHT CET 2019 (3 May, Shift 1)

Q49MHT CET 2019 · 3 MAY · SHIFT 1Medium

Hydrogen Spectrum & Spectral Series

In hydrogen emission spectrum, for any series, the principal quantum number is $n$ . Corresponding maximum wavelength $\lambda$ is ( $R =$ Rydberg's constant)

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Answer: (b)

Maximum wavelength of a series ending on level $n$ comes from $(n+1) \to n$ .
$\frac{1}{\lambda} = R\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right) = \frac{R(2n+1)}{n^2(n+1)^2}$
$\lambda = \frac{n^2(n+1)^2}{R(2n+1)}$
So the answer is (b).

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Hydrogen Spectrum & Spectral Series question from MHT CET 2019 (3 May, Shift 1)

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