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What is the principal quantum no. of H-atom Orbital if the electron energy is -3.4 eV ? also report the angular momentum of electron


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$E_1 for H = -13.6 eV$
$E_n = \large\frac{E_1}{E_2}$
= -3.4 = $\large\frac{-13.6}{n^2}$
n = 2
Now Angular momentum (mur) = $\large\frac{n.h}{2\pi}$
$= \large\frac{2\times6.626\times10^{-34}} {2\times3.14}$
Hence answer is (a)
answered Jan 31, 2014 by sharmaaparna1

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