We are asked to find the radius of a \( \text{Li}^{2+} \) ion (hydrogen-like ion with atomic number \( Z=3 \)) in its ground state in terms of the Bohr radius \( a_0 \).
According to Bohr’s atomic model, the radius of the \( n^{\text{th}} \) orbit for a hydrogen-like ion is given by:
\[ r_n = \frac{n^2 a_0}{Z}. \]
Here, \( a_0 = 0.529 \, \text{Å} \) is the Bohr radius for hydrogen (\( Z=1 \)).
Step 1: For \( \text{Li}^{2+} \), the atomic number is \( Z = 3 \).
Step 2: For the ground state, \( n = 1 \).
Step 3: Substitute into the formula:
\[ r_1 = \frac{1^2 \, a_0}{3} = \frac{a_0}{3}. \]
\[ r = \frac{a_0}{3} = \frac{1}{X}a_0. \] \[ \therefore X = 3. \]
Answer: \( X = 3 \)

Let \( \alpha = \dfrac{-1 + i\sqrt{3}}{2} \) and \( \beta = \dfrac{-1 - i\sqrt{3}}{2} \), where \( i = \sqrt{-1} \). If
\[ (7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}, \] then the value of \( m \) is ___________.