Question

The species having identical radii according to the Bohr's theory are:
A. H (first orbit)
B. \( He^+ \) (first orbit)
C. \( He^+ \) (Second orbit)
D. \( Li^{2+} \) (first orbit)
E. \( Be^{3+} \) (Second orbit)
Choose the correct answer from the options given below:

• A. A and C Only
• B. A and E Only
• C. B and E Only
• D. C and D Only

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
According to Bohr's model for hydrogen-like (single-electron) species, the radius of an orbit depends on the principal quantum number $n$ and the atomic number $Z$. Species with the same ratio of $n^2 / Z$ will have identical orbital radii.
Step 2: Key Formula or Approach:
Bohr radius formula: $r = 0.529 \frac{n^2}{Z} \text{ \AA}$
Therefore, $r \propto \frac{n^2}{Z}$.
Step 3: Detailed Explanation:
Let's calculate the proportionality factor $\frac{n^2}{Z}$ for each given species:
A. H (first orbit): $Z = 1$, $n = 1$
$\frac{n^2}{Z} = \frac{1^2}{1} = 1$
B. He$^+$ (first orbit): $Z = 2$, $n = 1$
$\frac{n^2}{Z} = \frac{1^2}{2} = 0.5$
C. He$^+$ (Second orbit): $Z = 2$, $n = 2$
$\frac{n^2}{Z} = \frac{2^2}{2} = \frac{4}{2} = 2$
D. Li$^{2+$ (first orbit)}: $Z = 3$, $n = 1$
$\frac{n^2}{Z} = \frac{1^2}{3} = \frac{1}{3} \approx 0.33$
E. Be$^{3+$ (Second orbit)}: $Z = 4$, $n = 2$
$\frac{n^2}{Z} = \frac{2^2}{4} = \frac{4}{4} = 1$
Comparing the calculated values:
Species A (H) and E (Be$^{3+}$) both yield $\frac{n^2}{Z} = 1$. Thus, their orbital radii are identical.
Step 4: Final Answer:
Options A and E have identical radii.