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H-atoms in ground state (13.6eV) are excited by monochromatic radiations of photon of energy 12.1eV. Find the number of spectral lines emitted in H atom.


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$\therefore n^2=\large\frac{E_1}{E_n}$
$\Rightarrow \large\frac{13.6}{1.5}$$=9$
$\Rightarrow n=3$
Thus transition occurs from $3^{rd}$ to $1^{st}$ orbit.
$\therefore$ spectral lines emitted $=\sum \Delta n$
$\Rightarrow \sum(3-1)$
$\Rightarrow \sum(2)$
$\Rightarrow 1+2=3$
Hence (c) is the correct answer.
answered Jan 17, 2014 by sreemathi.v

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