Question

The radius of the first orbit of \(Li^{2+}\) is \(X\ \text{\AA}\). The radius of the third orbit of \(He^{+}\) (in ) is

• A. \(\dfrac{18}{2}X\)
• B. \(\dfrac{18}{3}X\)
• C. \(\dfrac{27}{4}X\)
• D. \(\dfrac{27}{2}X\)

Hint:

For hydrogen-like species, \[ r_n \propto \frac{n^2}{Z} \] Always compare radii using the relation \(r_n=\dfrac{n^2a_0}{Z}\).

Answer 1

The Correct Option is D

Step 1: Write the Bohr radius as a ratio.
For any hydrogen-like ion, $r_n=\dfrac{n^2 a_0}{Z}$. Dividing two radii directly lets $a_0$ cancel out, so there is no need to solve for it first.

Step 2: Set up the ratio between the two orbits.
\[ \frac{r_3(He^+)}{r_1(Li^{2+})}=\frac{3^2/2}{1^2/3}=\frac{9/2}{1/3} \]

Step 3: Simplify.
\[ \frac{9/2}{1/3}=\frac{9}{2}\times 3=\frac{27}{2} \]

Step 4: Use the given value of $X$.
Since $r_1(Li^{2+})=X$,
\[ r_3(He^+)=\frac{27}{2}X \]
\[ \boxed{\frac{27}{2}X} \]