Question

The Bohr radius of a hydrogen like species is \(70.53\,\text{pm}\). The species and the stationary state \((n)\) are respectively. (Given: Hydrogen atom Bohr radius is \(52.9\,\text{pm}\)).

• A. \(Li^{2+},\,3\)
• B. \(He^{+},\,3\)
• C. \(He^{+},\,2\)
• D. \(Li^{2+},\,2\)

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to identify the atomic number \(Z\) and principal quantum number \(n\) that yield a specific radius using the Bohr model.
Step 2: Key Formula or Approach:
Radius of \(n\)-th orbit of a hydrogen-like species:
\[ r_n = a_0 \frac{n^2}{Z} \]
where \(a_0 = 52.9 \text{ pm}\).
Step 3: Detailed Explanation:
Given \(r_n = 70.53 \text{ pm}\).
\[ 70.53 = 52.9 \times \frac{n^2}{Z} \]
\[ \frac{n^2}{Z} = \frac{70.53}{52.9} \approx 1.333 = \frac{4}{3} \]
Comparing this ratio with the options:
- For \(\text{Li}^{2+}\) (\(Z=3\)) in \(n=2\) state: \(\frac{n^2}{Z} = \frac{2^2}{3} = \frac{4}{3}\). Correct.
- For \(\text{Li}^{2+}\) (\(Z=3\)) in \(n=3\) state: \(\frac{3^2}{3} = 3 \neq 1.33\).
- For \(\text{He}^{+}\) (\(Z=2\)) in \(n=2\) state: \(\frac{2^2}{2} = 2 \neq 1.33\).
Step 4: Final Answer:
The species is \(\text{Li}^{2+}\) and the state is \(n=2\).