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The ionization energy of $He^+$ is $19.6\times 10^{-18}J/atom$.Calculate the energy of first stationary state of $Li^{2+}$

$\begin{array}{1 1}(a)\;8.3\times 10^{-10}J/atom&(b)\;9.4\times 10^{-12}J/atom\\(c)\;44.1\times 10^{-18}J/atom&(d)\;44.1\times 10^{-12}J/atom\end{array}$

1 Answer

$E_1$ for $Li^{2+}=E_1$ for $H\times 9$
$E_1$ for $He^{+}=E_1$ for $H\times 4$
$E_1$ for $Li^{2+}=E_1$ for $He^+\times \large\frac{9}{4}$
$\Rightarrow 19.6\times 10^{-18}\times {9}{4}$
$\Rightarrow 44.1\times 10^{-18}j/atom$
Hence (c) is the correct answer.
answered Jan 15, 2014 by sreemathi.v

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