Step 1: Understanding the Concept:
Bohr's radius (\(a_0\)) specifically refers to the most probable distance between the nucleus and the electron in the ground state (\(n=1\)) of a hydrogen atom.
As the electron moves to higher energy levels (higher \(n\)), the centrifugal effect and quantization rules dictate that it must orbit at larger distances.
Step 2: Key Formula or Approach:
The mathematical formula for the radius of the \(n^{th}\) orbit in a Bohr atom is:
\[ r_n = \frac{n^2 h^2 \epsilon_0}{\pi m e^2} \]
For hydrogen (\(Z=1\)), we can simplify this as:
\[ r_n = n^2 \times a_0 \]
Where \(a_0 = \frac{h^2 \epsilon_0}{\pi m e^2} \approx 0.529 \text{ \AA}\).
Step 3: Detailed Explanation:
The question asks specifically for the radius of the "2nd Bohr orbit".
This means we set the principal quantum number \(n = 2\).
Substitute \(n = 2\) into our simplified relation:
\[ r_2 = (2)^2 \times a_0 \]
\[ r_2 = 4 \times a_0 = 4a_0 \]
This shows that the radius is not directly proportional to \(n\), but rather to its square.
Consequently, the second shell is much further away than twice the distance of the first shell; it is actually four times further.
For context:
Shell 1 (\(n=1\)) \(\rightarrow r = a_0\).
Shell 2 (\(n=2\)) \(\rightarrow r = 4a_0\).
Shell 3 (\(n=3\)) \(\rightarrow r = 9a_0\).
Step 4: Final Answer:
The radius of the 2nd Bohr orbit of Hydrogen is \(4a_0\).