The Bohr radius of a hydrogen like species is 70.53 pm. The species...

The radius of the $n$-th Bohr orbit for a hydrogen-like species is given by the formula:

$r_n = a_0 \dfrac{n^2}{Z}$

where $a_0$ is the Bohr radius of the hydrogen atom ($52.9$ pm), $n$ is the principal quantum number, and $Z$ is the atomic number.

Given $r_n = 70.53$ pm, we can write:

$70.53 = 52.9 \times \dfrac{n^2}{Z}$

$\dfrac{n^2}{Z} = \dfrac{70.53}{52.9} = 1.333 = \dfrac{4}{3}$

Checking the given options:
For $\text{Li}^{2+}$ and $n = 3$, $Z = 3 \Rightarrow \dfrac{n^2}{Z} = \dfrac{9}{3} = 3$
For $\text{He}^+$ and $n = 3$, $Z = 2 \Rightarrow \dfrac{n^2}{Z} = \dfrac{9}{2} = 4.5$
For $\text{He}^+$ and $n = 2$, $Z = 2 \Rightarrow \dfrac{n^2}{Z} = \dfrac{4}{2} = 2$
For $\text{Li}^{2+}$ and $n = 2$, $Z = 3 \Rightarrow \dfrac{n^2}{Z} = \dfrac{4}{3}$

The ratio matches for $\text{Li}^{2+}$ and $n = 2$.

Answer: $\text{Li}^{2+}, 2$