Question

The radius of third orbit of He\(^+\) is \( Y \, \text{Å} \). The radius of second orbit of Li\(^{2+}\) (in Å) is:

• A. \( \frac{27}{4} Y \)
• B. \( \frac{27}{8} Y \)
• C. \( \frac{8}{27} Y \)
• D. \( \frac{4}{27} Y \)

Hint:

For hydrogen-like ions, radius scales as \( r \propto \frac{n^2}{Z} \). Always compare using ratios to avoid full calculation.

Answer 1

The Correct Option is C

Step 1: Use Bohr’s formula for hydrogen-like species.
For hydrogen-like ions, the radius of nth orbit is given by: \[ r_n = \frac{a_0 n^2}{Z} \] where \( a_0 \) is Bohr radius, \( n \) is orbit number, and \( Z \) is atomic number.

Step 2: Express radius of He\(^+\) third orbit.
For He\(^+\), \( Z = 2 \), \( n = 3 \): \[ Y = \frac{a_0 \cdot 9}{2} \] So, \[ a_0 = \frac{2Y}{9} \]

Step 3: Find radius of Li\(^{2+}\) second orbit.
For Li\(^{2+}\), \( Z = 3 \), \( n = 2 \): \[ r = \frac{a_0 \cdot 4}{3} \] Substitute \( a_0 = \frac{2Y}{9} \): \[ r = \frac{4}{3} \cdot \frac{2Y}{9} \]

Step 4: Simplify the expression.
\[ r = \frac{8Y}{27} \]

Step 5: Physical interpretation.
Higher nuclear charge \( Z \) reduces orbital radius significantly, hence Li\(^{2+}\) has a much smaller radius compared to He\(^+\) scaled value.
Final Answer: \[ \boxed{\frac{8}{27} Y} \]