Question

The radius of the n^th orbit of a single electron species is $0.132{\mathrm{n}}^2{\mathrm{A}}^{\circ }$. Identify the element.

Answer 1

1. The radius of a single electron species' n^th orbit = $0.132{\mathrm{n}}^2{\mathrm{A}}^{\circ }$.
2. The radius of n^th orbit of a single electron species is given as; $r_n=0.529\left(\frac{n^2}{Z}\right)A^{°}$; Where $Z$= atomic number of single-electron species.
3. Thus,

$$\begin{gathered} 0.132n^2=0.529\left(\frac{n^2}{Z}\right)A^{°} \\ \Rightarrow Z=4 \end{gathered}$$

4. Since the atomic number $\left(Z\right)=4$.

5. Thus, the element is beryllium ($\mathrm{Be}$).