Question

Bohr's radius of 2nd orbit of $ \text{Be}^{3+} $ is equal to that of

• A. 4th orbit of hydrogen
• B. 2nd orbit of He\(^{+} \)
• C. 3rd orbit of Li\(^{2+} \)
• D. 1st orbit of hydrogen

Hint:

The Bohr radius depends on the atomic number \( Z \) and the quantum number \( n \). For the same quantum number \( n \), a larger \( Z \) leads to a smaller orbit.

Answer 1

The Correct Option is D

The radius of an electron's orbit in Bohr's model is given by the formula: \[ r_n = \frac{n^2 h^2 \epsilon_0}{\pi m e^2 Z} \] Where: 
- \( n \) is the principal quantum number, 
- \( h \) is Planck's constant, 
- \( \epsilon_0 \) is the permittivity of free space, 
- \( m \) is the mass of the electron, 
- \( e \) is the charge of the electron, and 
- \( Z \) is the atomic number of the ion. We are asked to compare the Bohr radius of the 2nd orbit of \( \text{Be}^{3+} \) and various orbits of other atoms. For \( \text{Be}^{3+} \), the atomic number is \( Z = 4 \), and for the second orbit, \( n = 2 \). For hydrogen, \( Z = 1 \), and for the first orbit, \( n = 1 \). 

The formula shows that the radius of the orbit is inversely proportional to \( Z \), and directly proportional to \( n^2 \). 
Thus, the Bohr radius for the second orbit of \( \text{Be}^{3+} \) can be compared with the first orbit of hydrogen, as they both satisfy the relationship of proportionality when adjusted for their respective values of \( n \) and \( Z \). 

Therefore, the Bohr radius of the 2nd orbit of \( \text{Be}^{3+} \) is equal to the Bohr radius of the 1st orbit of hydrogen. 
Thus, the correct answer is 1st orbit of hydrogen.