Q148 / 150 · MHT CET 2026 (18 April Shift 1) · Physics
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$n = 3$
- Correct answerC$n = 2$
- D$n = 1$
Solution
The radius of the $n$-th Bohr orbit for a hydrogen-like atom is given by $r = r_0 \dfrac{n^2}{Z}$, where $r_0$ is the Bohr radius.
For the ground state of hydrogen, $n = 1$ and $Z = 1$.
$r_H = r_0 \dfrac{1^2}{1} = r_0$
For triply ionized beryllium ($\text{Be}^{+++}$), $Z = 4$. Let its principal quantum number be $n$.
$r_{Be} = r_0 \dfrac{n^2}{4}$
Given that the orbital radii are equal, $r_{Be} = r_H$.
$r_0 \dfrac{n^2}{4} = r_0$
$n^2 = 4$
$n = 2$
Answer: $n = 2$
For the ground state of hydrogen, $n = 1$ and $Z = 1$.
$r_H = r_0 \dfrac{1^2}{1} = r_0$
For triply ionized beryllium ($\text{Be}^{+++}$), $Z = 4$. Let its principal quantum number be $n$.
$r_{Be} = r_0 \dfrac{n^2}{4}$
Given that the orbital radii are equal, $r_{Be} = r_H$.
$r_0 \dfrac{n^2}{4} = r_0$
$n^2 = 4$
$n = 2$
Answer: $n = 2$
Hence, the correct answer is option: 'C'.