Question

The difference in $n^{th}$ and $(n+1{)}^{th}$ Bohr's radius of $H$-atoms is equal to $(n-1{)}^{th}$ Bohr's radius. Hence the value of ${}^{\prime }{n}^{\prime }$ is :

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is D $4$

By using the equation:
$r=\frac{0.529\times n^2}{1}\stackrel{o}{A}$
$\therefore 0.529\left[\right(n+1{)}^2-n^2]=0.529(n-1{)}^2\Rightarrow n^2+2n+1-n^2=n^2-2n+1\Rightarrow n^2-4n=0\Rightarrow n(n-4)=0\Rightarrow n=4\text{or}n=0$
The value of $n$ can't be zero.
So, $n=4$