First slide

Quantum mechanical model of the atom

Question

In Lyman series, shortest wavelength of H-atom appears at x meter , then longest wavelength in Balmer series of He⁺ appear at

Moderate

A

$\frac{9 x}{5} m$
B

$\frac{36 x}{5} m$
C

$\frac{x}{4} m$
D

$\frac{5 x}{9}$

Solution

For the spectral line in H-atom and H-like species (one electron)
$\frac{1}{λ} = \bar{v} = {\bar{R}}_{H} Z^{2} [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$
For Lyman series
For shortest wavelength (maximum wave number) $n_{2} \to \infty$
$∴ \frac{1}{λ_{\min}} = v_{\max} = {\bar{R}}_{H} {(1)}^{2} [\frac{1}{1}]$
For longest wavelength (minimum wave number), $n_{2} = (n_{1} + 1)$
For Balmer series, $n_{1} = 2$
$∴ \frac{1}{λ_{\max}} = v_{\min} = R_{H} {(Z)}^{2} [\frac{1}{2}]$
For He⁺ , Z = 2
$\frac{1}{λ_{\max}} = \frac{1}{x} {(2)}^{2} [\frac{5}{36}] \Rightarrow \frac{1}{λ_{\max}} = \frac{5}{9 x} ∴ λ_{\max} = \frac{9 x}{5}$

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