If the shortest wavelength of the spectral line of H - atom in Lyman series is x, then the longest wavelength of the line in Balmer series of $L i^{2 +}$ is

$\frac{5 x}{4}$

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$\frac{4 x}{5}$

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$\frac{x}{9}$

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$9 x$

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Solution

The correct option is B
$\frac{4 x}{5}$

$v_{m a x} = \frac{1}{λ_{m i n}} = R_{H} \frac{Z^{2}}{n_{1}^{2}} (n_{2} = \infty)$
For H - atom, Z = 1 : For Lyman series, $n_{1} = 1$
Hence, $\frac{1}{x} = R_{H}$
For largest wavelength in the Balmerseries of $L i^{2 +}, Z = 3, n_{1} = 2; n_{2} = 3$
Hence, $\frac{1}{λ_{m a x^{'}}} R_{H} Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
$= \frac{1}{x} \times 3^{2} (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = \frac{9}{x} [\frac{5}{36}] = \frac{5}{4 x}$
or $λ_{m a x} = \frac{4 x}{5}$

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If the shortest wavelength of the spectral line of H - atom in Lyman series is x, then the longest wavelength of the line in Balmer series of Li2+ is

The correct option is B 4x5 vmax=1λmin=RHZ2n21(n2=∞) For H - atom, Z = 1 : For Lyman series, n1=1 Hence, 1x=RH For largest wavelength in the Balmerseries of Li2+,Z=3,n1=2;n2=3 Hence, 1λmax′RHZ2(1n21−1n22) =1x×32(122−132)=9x[536]=54x or λmax=4x5

If the shortest wavelength of the spectral line of H - atom in Lyman series is x, then the longest wavelength of the line in Balmer series of $L i^{2 +}$ is