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The radius of which of the following orbit is same as that of the first Bohr's orbit of hydrogen atom

$\begin{array}{1 1}(a)\;He^+(n=2)&(b\;Li^{2+}(n=2)\\(c)\;Li^{2+}(n=3)&(d)\;Be^{3+}(n=2)\end{array}$

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$r_2Be^{3+}=\large\frac{r_2H}{z}$
$\Rightarrow \large\frac{r_1H\times 2^2}{z}$
$\Rightarrow \large\frac{r_1H\times 4}{4}$
$\Rightarrow r_1H$
Hence (d) is the correct answer.
answered Jan 15, 2014 by sreemathi.v
 

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