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If the electron energy is -3.4 eV , Find the principal quantum number of H-atom?


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$E_n = \large\frac{E_1}{E_2}$
(Since $E_1$ for H-atom = -13.6 eV
Or $n^2 = \large\frac{E_1}{E_2}$
$= \large\frac{-13.6}{-3.4}=4$
$\therefore$ n = 2
Hence the answer is (c)


answered Jan 24, 2014 by sharmaaparna1
edited Jan 24, 2014 by sharmaaparna1

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