In spectrum of hydrogen atom, ratio of longest wavelength in Lyman series to thelongest wavelength i

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12.Atoms

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In the spectrum of hydrogen atom, the ratio of the longest wavelength in Lyman series to thelongest wavelength in the Balmer series is

A

$5/27$

B

$1/93$

C

$4/9$

D

$3/2$

Solution

In Lyman series $\lambda_{\max }=\frac{4}{3 \mathrm{R}}$

In Balmer series $\lambda_{\max }=\frac{36}{5 \mathrm{R}}$

So required ratio $=\frac{5}{27}$

Standard 12

Physics

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