Question

Which energy state of triply ionized Beryllium ${\text{Be}}^{3+}$ has the same orbital radius as that of ground state of hydrogen atom?

• A. $n=8\text{state}$
• B. $n=5\text{state}$
• C. $n=4\text{state}$
• D. $n=2\text{state}$

Answer 1

The correct option is D $n=2\text{state}$

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

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

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

Let $(r_1,n_1,\text{and}{Z}_1)\&(r_2,n_2,\text{and}{Z}_2)$ be the radius of the orbit, energy state and atomic no for hydrogen atom and ${\text{Be}}^{3+}$ respectively.

Here, $Z_1=1;Z_2=4$

$\Rightarrow \frac{r_2}{r_1}=\left(\frac{n_2^2}{n_1^2}\right)\times \left(\frac{Z_1}{Z_2}\right)$

$[\because r_1=r_2;n_1=1]$

$\Rightarrow n_2^2=n_1^2\times \left(\frac{Z_2}{Z_1}\right)=1^2\times \left(\frac{4}{1}\right)$

$\Rightarrow n_2^2=4$

$\therefore n_2=2$

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

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