Determine Hydrogen Like Species and Quantum State from Bohr Radius

The radius of the $n^{\text{th}}$ stationary orbit for a hydrogen-like species with atomic number $Z$ is given by Bohr's formula:

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

where:

• $r_n$ is the radius of the orbit,
• $a_0 = 52.9\text{ pm}$ is the Bohr radius of the hydrogen atom ($n=1, Z=1$),
• $n$ is the principal quantum number (stationary state),
• $Z$ is the atomic number of the species.

Given $r_n = 70.53\text{ pm}$ and $a_0 = 52.9\text{ pm}$, we can set up the ratio:
$\frac{n^2}{Z} = \frac{r_n}{a_0}$

Substituting the given values into the equation:
$\frac{n^2}{Z} = \frac{70.53\text{ pm}}{52.9\text{ pm}} \approx 1.3333 = \frac{4}{3}$

Now, let's evaluate $\frac{n^2}{Z}$ for each option:

1. Option A: $\text{Li}^{2+}$ ($Z = 3$), $n = 3$ $\frac{n^2}{Z} = \frac{3^2}{3} = 3$

2. Option B: $\text{He}^{+}$ ($Z = 2$), $n = 3$ $\frac{n^2}{Z} = \frac{3^2}{2} = 4.5$

3. Option C: $\text{He}^{+}$ ($Z = 2$), $n = 2$ $\frac{n^2}{Z} = \frac{2^2}{2} = 2$

4. Option D: $\text{Li}^{2+}$ ($Z = 3$), $n = 2$ $\frac{n^2}{Z} = \frac{2^2}{3} = \frac{4}{3} \approx 1.3333$

Thus, the species is $\text{Li}^{2+}$ and the stationary state is $n = 2$.

Correct Option: D ($\text{Li}^{2+}, 2$)