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Given that, $r_{(n+1)}-r_n=r_{(n-1)}$,where $r_n,r_{n-1}$ and $r_{(n+1)}$ are Bohr's radius for H-atom in $n^{th},(n-1)^{th}$ and $(n+1)^{th}$ shell respectively.

$(a)\;3\qquad(b)\;2\qquad(c)\;4\qquad(d)\;5$

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$r_n=n^2\times r_1$
$\Rightarrow r_{n+1}=(n+1)^2\times r_1$
$\Rightarrow r_{n-1}=(n-1)^2\times r_1$
$\Rightarrow r_{(n+1)}-r_{n}=r_{(n-1)}$
$\therefore (n+1)^2r_1-n^2r_1=(n-1)^2r_1$
$\Rightarrow (n^2+2n+1)-n^2=(n^2-2n+1)$
$\Rightarrow n^2-4n=0$
$\therefore n=4$
Hence (c) is the correct answer.
answered Jan 17, 2014 by sreemathi.v
edited Mar 21, 2014 by mosymeow_1
 

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