(9am to 6pm)

Ask Questions, Get Answers

Want help in doing your homework? We will solve it for you. Click to know more.
Home  >>  JEEMAIN and NEET  >>  Chemistry  >>  Atomic Structure

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.


1 Answer

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

Related questions