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The wavelength of $m^{th}$ line Balmer series for an orbital is $4103 A^{\large\circ}$. What is the value of $m$?


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For Balmer Series
$\Rightarrow\large\frac {1} {4103\times10^{-8} }$= $109678[\large\frac{1}{2^2}-\large\frac{1}{(m+2)^2}]$
$\Rightarrow [\large\frac{1}{2^2}-\frac{1}{(m+2)^2}] = \frac{10^8}{4103 \times 109678}$
$\Rightarrow \large\frac{1}{(m+2)^2} = 0.02778$
$\Rightarrow\large\frac{1}{(m+2)} = 0.1667$
$\Rightarrow m+2 =5.99 \approx 6$
$\therefore m = 4$
$m^{th}$ line represents $4^{th}$ line of balmer series
Hence answer is (c)
answered Jan 23, 2014 by sharmaaparna1
edited Mar 21, 2014 by mosymeow_1

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