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Chemistry
The first emission line in the atomic spectrum of hydrogen in the Balmer series appears at
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Q. The first emission line in the atomic spectrum of hydrogen in the Balmer series appears at
Structure of Atom
A
$\frac{9R}{400} cm^{-1}$
13%
B
$\frac{7R}{144} cm^{-1}$
14%
C
$\frac{3R}{4} cm^{-1}$
23%
D
$\frac{5R}{36} cm^{-1}$
50%
Solution:
For Balmer $n_1 = 2$ and $n_2 = 3; v$
$= R \left(\frac{1}{2^{2}} - \frac{1}{3^{2}}\right) = \frac{5R}{36} cm^{-1}$