Question

The triply ionized beryllium (\(\text{Be}^{+++}\)) has the same electron orbital radius as that of the ground state of hydrogen. Hence, the energy state of triply ionized beryllium is (Given \(Z = 4\) for beryllium)

• A. \(n = 4\)
• B. \(n = 3\)
• C. \(n = 2\)
• D. \(n = 1\)

Hint:

Bohr radius is proportional to n squared over Z.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The radius of the \(n\)th Bohr orbit of a hydrogen-like ion is \(r_n = \frac{n^2a_0}{Z}\), where \(a_0\) is the radius of the first orbit of hydrogen.

Step 2: Key Formula or Approach:
The ground state of hydrogen has \(r = a_0\). For \(\text{Be}^{3+}\), \(Z = 4\).

Step 3: Detailed Explanation:
Set \(\frac{n^2a_0}{4} = a_0\).
\(n^2 = 4\), so \(n = 2\).
\[ n = 2 \]
The state \(n = 1\) would have a radius of \(\frac{a_0}{4}\), and \(n = 3\) would have a radius of \(\frac{9a_0}{4}\).

Final Answer:
The energy state is \(n = 2\), option (C). \[ \boxed{n = 2} \]