Question

Difference between $n^{th}$ and $(n+2{)}^{th}$ Bohr's radius of $H$ atom is equal to its $(n-2{)}^{th}$ Bohr's radius. The value of $n$ is:

• A. 4
• B. 5
• C. 6
• D. 8

Answer 1

The correct option is D 8

As per the question, for $H$ atom
$r_{n+2}-r_n=r_{n-2}...\left(i\right)$
where $r_n=\text{Radius of Bohr's nth energy level}$
As per Bohr's theory, $r_n=0.529\frac{n^2}{Z}$
For $H$ atom, $r_n=0.529n^2$
Similiarly, $r_{n+2}=0.529(n+2{)}^2$
$r_{n-2}=0.529(n-2{)}^2$
Putting these values in equation (i), we get
$0.529(n+2{)}^2-0.529n^2=0.529(n-2{)}^2$
$\Rightarrow (n+2{)}^2-n^2=(n-2{)}^2$
$\Rightarrow n^2-8n=0$
$\Rightarrow n(n-8)=0$
$\Rightarrow n=0\text{or}n=8$
Since $n=0$ is not possible, the only possible value is $n=8$.