JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Modern Physics

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Which of the following statement(s) is(are) correct about the spectrum of hydrogen atom?
(A) The ratio of the longest wavelength to the shortest wavelength in Balmer series is 9/5
(B) There is an overlap between the wavelength ranges of Balmer and Paschen series.
(C) The wavelengths of Lyman series are given by

(1 + \frac{1}{m ^{2}}) λ_{0}

, where

λ_{0}

is the shortest wavelength of Lyman series and m is an integer
(D) The wavelength ranges of Lyman and Balmer series do not overlap

(A)The ratio of the longest wavelength to the shortest wavelength in Balmer series is 9/5✓ Correct
(B)There is an overlap between the wavelength ranges of Balmer and Paschen series.
(C)The wavelengths of Lyman series are given by $(1 + \frac{1}{m ^{2}}) λ_{0}$ , where $λ_{0}$ is the shortest wavelength of Lyman series and m is an integer
(D)The wavelength ranges of Lyman and Balmer series do not overlap✓ Correct

Answer: (A), (D)

Solution

For (A): Balmer series has transitions to n = 2. Longest wavelength:

n = 3 \to 2

, giving

\frac{1}{λ _{m a x}} = R [\frac{1}{4} - \frac{1}{9}] = \frac{5 R}{36}

. Shortest wavelength:

n = \infty \to 2

, giving

\frac{1}{λ _{m i n}} = R [\frac{1}{4}] = \frac{R}{4}

. Ratio:

\frac{λ _{m a x}}{λ _{m i n}} = \frac{36/5 R}{4/ R} = \frac{9}{5}

. ✓

For (B): Longest wavelength of Balmer =

\frac{36}{5 R}

. Shortest wavelength of Paschen =

\frac{9}{R}

. Since

\frac{36}{5 R} < \frac{9}{R}

, there is no overlap. ✗

For (C): For Lyman series:

\frac{1}{λ} = R [1 - \frac{1}{m ^{2}}]

, so

λ = \frac{λ _{0}}{1 - 1/ m ^{2}} = \frac{λ _{0} m ^{2}}{m ^{2} - 1}

, not

(1 + 1/ m^{2}) λ_{0}

. ✗

For (D): Longest wavelength of Lyman =

\frac{4}{3 R}

. Shortest wavelength of Balmer =

\frac{4}{R}

. Since

\frac{4}{3 R} < \frac{4}{R}

, there is no overlap. ✓

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JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Modern Physics

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