Which of the following statement(s) is (are) correct about the spectrum of hydrogen atom? [2021]

Question

Which of the following statement(s) is (are) correct about the spectrum of hydrogen atom? [2021]

Smart Achievers · Accepted Answer

(1, 4)

From formula,

(1) $\frac{1}{λ} = R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$

For Balmer series, $n_{1} = 2$

$∴$ For longest wavelength, transition occurs from $n = 3$ to $n = 2$

$∴$ $\frac{1}{λ_{\max}} = R [\frac{1}{2^{2}} - \frac{1}{3^{2}}]$

For shortest wavelength, transition occurs from $n = \infty$ to $n = 2$

$∴$ $\frac{1}{λ_{\min}} = R [\frac{1}{2^{2}} - \frac{1}{\infty^{2}}]$

$∴$ $\frac{λ_{longest}}{λ_{shortest}} = \frac{9}{5}$

(2) For Paschen series, $n_{1} = 3$

$λ_{longest of Balmer} = \frac{36}{5 R}$ and

$λ_{shortest of Paschen} = \frac{9}{R}$

Hence there is no overlap between the wavelength ranges of Balmer and Paschen series.

(3), (4) For Lyman series, $n_{1} = 1$

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

Also, $\frac{1}{λ_{0}} = R$

$∴$ $\frac{1}{λ} = \frac{1}{λ_{0}} [1 - \frac{1}{m^{2}}]$ $\Rightarrow λ = \frac{λ_{0}}{1 - \frac{1}{m^{2}}}$

$λ_{longest of Lyman} = \frac{4}{3 R}$ and $λ_{shortest of Balmer} = \frac{4}{R}$

Hence the wavelength ranges of Lyman and Balmer series do not overlap.

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