Physics - Modern Physics Question with Solution | TestHub

The radius of the $n$-th Bohr orbit for a hydrogen-like atom is given by $r=r_0\frac{n^2}{Z}$, where $r_0$ is the Bohr radius.

For the ground state of hydrogen, $n=1$ and $Z=1$.

$r_H=r_0\frac{1^2}{1}=r_0$

For triply ionized beryllium (${\text{Be}}^{+++}$), $Z=4$. Let its principal quantum number be $n$.

$r_{Be}=r_0\frac{n^2}{4}$

Given that the orbital radii are equal, $r_{Be}=r_H$.

$r_0\frac{n^2}{4}=r_0$

$n^2=4$

$n=2$

Answer: $n=2$