Question

Determine the ratio of perimeter of $2^{nd}$ and $3^{rd}$ Bohr orbit in ${\text{He}}^{+}$ atom $(Z=2)$.

• A. $\frac{9}{4}$
• B. $\frac{16}{9}$
• C. $\frac{9}{16}$
• D. $\frac{4}{9}$

Answer 1

The correct option is D $\frac{4}{9}$

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

$r_n=0.529\left(\frac{n^2}{Z}\right)\mathring{\text{A}}$

$\Rightarrow r_n\propto \frac{n^2}{Z}$

$\Rightarrow \frac{r_2}{r_3}={\left(\frac{n_2}{n_3}\right)}^2={\left(\frac{2}{3}\right)}^2=\frac{4}{9}[\because Z=\text{same}]$

Now, perimeter of the $n^{th}$ bohr radius is,

$P_n=2\pi r_n\Rightarrow P_n\propto r_n$

Let $P_2$ and $P_3$ be the perimeter of the $2^{nd}$ and $3^{rd}$ bohr orbit, respectively.

$\Rightarrow \frac{P_2}{P_3}=\frac{r_2}{r_3}=\frac{4}{9}$

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

Why this question ? Key concept : The bohr radius of the $n^{th}$ orbit is,$r_n=0.53\left(\frac{n^2}{Z}\right)\mathring{A}$ $\therefore r_n\propto \frac{n^2}{Z}$