The radius of the second Bohr orbit, in terms of the Bohr radius a0, in Li2+ is:

🧠 The Fundamental Unit Scaling We need the radius of the 2nd Bohr orbit for $\ce{Li^2+}$. In JEE, this is often asked "in terms of $a_0$" so we don't need to plug in $0.529$. Just manage $n$ and $Z$.

🗺️ The Variable Mapping

1. Formula: $r_n = \frac{n^2 a_0}{Z}$.
2. Ion: Lithium ($Z=3$).
3. State: 2nd orbit ($n=2$).
4. Result: $r_2 = \frac{2^2 a_0}{3} = \frac{4 a_0}{3}$

⚡ The "Squares and Atomic Numbers" Mental Table

• H, n=1 $\to$ $a_0$
• He+, n=1 $\to$ $a_0/2$
• Li2+, n=2 $\to$ $4/3 a_0$ Being able to visualize this simple table of $n^2/Z$ factors makes these questions trivial.

⚠️ Common Traps Wait—did you use $n=2$ or $n^2$? Squaring is the most missed step. $2/3 a_0$ is a common distractor option (Option A). Always double check the power of $n$!

$\boxed{\text{Answer: (c)}}$