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: Use Bohr's radius formula for hydrogen-like species.
For a hydrogen-like atom, \[ r_n=\frac{n^2a_0}{Z} \] where \(n\) is the principal quantum number, \(Z\) is the atomic number and \(a_0\) is the Bohr radius.

Step 2: Radius of the first orbit of \(Li^{2+}\).
For \(Li^{2+}\), \[ Z=3,\qquad n=1 \] Therefore, \[ X=r_1=\frac{1^2a_0}{3} =\frac{a_0}{3} \] Hence, \[ a_0=3X \]

Step 3: Radius of the third orbit of \(He^{+}\).
For \(He^{+}\), \[ Z=2,\qquad n=3 \] Thus, \[ r_3=\frac{3^2a_0}{2} =\frac{9a_0}{2} \] Substituting \(a_0=3X\), \[ r_3=\frac{9(3X)}{2} =\frac{27X}{2} \]

Step 4: Simplify the result.
Therefore, \[ r_3=\frac{27}{2}X \]

Step 5: Final conclusion.
Hence, \[ \boxed{\frac{27}{2}X} \]