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Estimate the wavelength of second line (also called as $\beta$-line) in Balmer series of hydrogen atom


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The transition responsible for second Balmer ($\beta$-line) line is $4\to 2$
$\Delta E=13.6(1)^2(\large\frac{1}{2^2}-\frac{1}{4^2}) = 2.55eV$
Now, $\lambda=\large\frac{hc}{\Delta E}$
$\Rightarrow \lambda=4.862\times 10^{-7}m$
$\Rightarrow \lambda=4862\;\overset{\circ}{A}$
Hence (c) is the correct answer.
answered Jan 9, 2014 by sreemathi.v
edited Mar 19, 2014 by mosymeow_1

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