Question

If the Bohr's radius of H-atom is 52.9 pm and radius of H-like species is 70.53 pm, then chemical species and orbit number are respectively.

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

Hint:

Bohr's radius for \( H \)-like species is directly related to the atomic number and the orbit number. For a hydrogen-like ion, the radius decreases as the atomic number increases.

Answer 1

The Correct Option is B

The Bohr radius for a hydrogen atom is given as 52.9 pm. The radius of an \( H \)-like species (i.e., an ion that is hydrogen-like) can be calculated using the formula:
\[ r = \frac{n^2}{Z} \times 52.9 \, \text{pm} \] Where:
- \( r \) is the radius of the \( H \)-like species,
- \( n \) is the orbit number (principal quantum number),
- \( Z \) is the atomic number of the species.
We are given that the radius of the \( H \)-like species is 70.53 pm. Substituting this into the formula: \[ 70.53 = \frac{n^2}{Z} \times 52.9 \] Now, solving for \( \frac{n^2}{Z} \): \[ \frac{n^2}{Z} = \frac{70.53}{52.9} = 1.33 \] Now, let's check the options: - For \( Li^{+2} \), \( Z = 3 \), so we calculate \( n^2 = 1.33 \times 3 = 4 \), which gives \( n = 2 \). - Thus, the chemical species is \( Li^{+2} \) and the orbit number is 2. Final Answer: Option (B) Li\(^{+2}\), 2.