Question

If the difference between $(n+1{)}^{th}$ Bohr radius and $n^{th}$ Bohr radius is equal to the $(n-1{)}^{th}$ Bohr radius, then the value of $n$ is

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

The correct option is A $4$

The Bohr radius of the $n^{th}$ orbit is,

$r_n=0.53\left(\frac{n^2}{Z}\right)\mathring{A}$

$\Rightarrow r_n\propto \left(\frac{n^2}{Z}\right)[\because Z=\text{same}]$

As, $n=1$, $r_n=r_0$

$\Rightarrow \frac{r_n}{r_0}\propto \left(\frac{n^2}{1^2}\right)$

$\Rightarrow r_n=r_0{n}^2$

Now, as per question,

$\Rightarrow \left[\right(n+1{)}^2-n^2]r_0=(n-1{)}^2{r}_0$

$\Rightarrow (n^2+2n+1-n^2)=(n^2-2n+1)$

$\Rightarrow 2n+1=n^2-2n+1$

$\Rightarrow n^2=4n$

$\Rightarrow n^2-4n=0$

$\Rightarrow n(n-4)=0$

$\Rightarrow n=0\text{(or)}n=4$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.

Why this question ? The bohr radius of the $n^{th}$ orbit is, $\because r_n=0.53\left(\frac{n^2}{Z}\right)\mathring{A}$ $\Rightarrow r_n\propto \left(\frac{n^2}{Z}\right)$ $\Rightarrow r_n=r_0{n}^2[\because r_0=0.53]$