Question

According to Bohr's theory, when principal quantum number is 2, the radius of \( \text{Be}^{3+} \) will be:

• A. 52.90 pm
• B. 26.45 pm
• C. 105.80 pm
• D. 13.23 pm

Hint:

For the radius of an electron in a hydrogen-like atom, Bohr's formula \( r_n = 0.529 \, \frac{n^2}{Z} \, \text{Å} \) is very useful, where \( n \) is the principal quantum number and \( Z \) is the atomic number.

Answer 1

The Correct Option is A

Bohr's formula for the radius of an electron in an atom is: \[ r_n = 0.529 \, \frac{n^2}{Z} \, \text{Å} \] where:
- \( n \) is the principal quantum number,
- \( Z \) is the atomic number.
For \( \text{Be}^{3+} \), \( Z = 4 \) and \( n = 2 \).
Substituting these values into the formula: \[ r_2 = 0.529 \times \frac{2^2}{4} = 0.529 \times \frac{4}{4} = 0.529 \, \text{Å} = 52.9 \, \text{pm} \] Thus, the radius of \( \text{Be}^{3+} \) when \( n = 2 \) is 52.90 pm.