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A spectral line in the spectrum of $H$-atom has a wave number of $15222.22cm^{-1}$. What transition is responsible for this radiation? (Rydberg constant $R=1.096\times 10^5cm^{-1}$)

$(a)\;3\qquad(b)\;1\qquad(c)\;2\qquad(d)\;4$

1 Answer

$\lambda=\large\frac{1}{\bar v}=\frac{1}{15222.22} = 6.569\times 10^{-5}cm = 6569\overset{\circ}{A}$
Clearly, the spectral line lies in the visible region (i.e.) in Balmer series
Hence, $n_1=2$
Using the relation for wave number for H-atom
$\Large\bar v=\frac{1}{\lambda}(\large\frac{1}{n_1^2}-\frac{1}{n_2^2})$
$\Rightarrow 15222.22=1.096\times 10^5\Large(\frac{1}{2^2}-\frac{1}{(n_2)^2})$
$\Rightarrow n_2=3$
Hence (a) is the correct answer.
answered Jan 9, 2014 by sreemathi.v
edited Mar 19, 2014 by mosymeow_1
 

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