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: Equate radii:
$\frac{n^2}{Z} = 1$ for the radius to equal $a_0$.

Step 2: Solve:
$n^2 = Z = 4$, so $n = 2$.

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