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Question

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


A

5x4

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B

4x5

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C

x9

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D

9x

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Solution

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, 1λmaxRHZ2(1n211n22)
=1x×32(122132)=9x[536]=54x
or λmax=4x5


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