Question:medium

The radius in the hydrogen atom in the ground state is $0.53$ \AA. The radius of $Li^{+2}$ ion (atomic number $= 3$) in a similar state is \ldots..

Show Hint

To quickly solve Bohr radius problems, remember that as the nuclear charge ($Z$) increases, the nucleus pulls the electron closer, making the radius smaller. Since Lithium has 3 protons ($Z=3$) compared to Hydrogen's 1, its ground state radius must be exactly one-third of Hydrogen's.
Updated On: Jun 3, 2026
  • $0.53 \text{\AA}$
  • $1.06 \text{\AA}$
  • $0.17 \text{\AA}$
  • $0.265 \text{\AA}$
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
Bohr's model explains that for hydrogen-like species (ions with only one electron), the orbit radius is dictated by the balance of electrostatics and centripetal force. Increasing the number of protons in the nucleus increases the attraction, pulling the electron closer.
Key Formula or Approach:
The radius \( r_n \) for the \( n \)-th orbit is given by:
\[ r_n = 0.529 \frac{n^2}{Z} \, \text{\AA} \]
where \( n \) is the principal quantum number and \( Z \) is the atomic number.
Step 2: Detailed Explanation:
For Hydrogen (H): \( Z = 1 \). Ground state means \( n = 1 \).
\( r_H = 0.529 \frac{1^2}{1} = 0.529 \text{\AA} \approx 0.53 \text{\AA} \).
For Lithium ion (\( \text{Li}^{2+} \)): \( Z = 3 \). "Similar state" implies ground state, so \( n = 1 \).
Using the ratio:
\[ \frac{r_{Li^{2+}}}{r_H} = \frac{Z_H}{Z_{Li^{2+}}} \]
\[ r_{Li^{2+}} = r_H \times \frac{1}{3} \]
\[ r_{Li^{2+}} = \frac{0.53}{3} \approx 0.1766 \text{\AA} \]
Looking at the options, \( 0.17 \text{\AA} \) is the closest approximation.
Step 3: Final Answer:
The radius of the \( \text{Li}^{2+} \) ion is \( 0.17 \text{\AA} \).
Was this answer helpful?
4