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Q)

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

$(a)\;2\qquad(b)\;3\qquad(c)\;4\qquad(d)\;5$

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A)
For Balmer Series
$\large\frac{1}{\lambda}$=$R_H[\large\frac{1}{n_1^2}-\large\frac{1}{n_2^2}]$
$\Rightarrow\large\frac {1} {4103\times10^{-8} }$= $109678[\large\frac{1}{2^2}-\large\frac{1}{(m+2)^2}]$
$\Rightarrow [\large\frac{1}{2^2}-\frac{1}{(m+2)^2}] = \frac{10^8}{4103 \times 109678}$
$\Rightarrow \large\frac{1}{(m+2)^2} = 0.02778$
$\Rightarrow\large\frac{1}{(m+2)} = 0.1667$
$\Rightarrow m+2 =5.99 \approx 6$
$\therefore m = 4$
$m^{th}$ line represents $4^{th}$ line of balmer series
Hence answer is (c)
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