Which state of triply ionised Beryllium (Be^{ | vedclass

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(C) The radius of the $n^{th}$ orbit for a hydrogen-like atom is given by the formula:$r_n = r_0 \left( \frac{n^2}{Z} \right)$where $r_0$ is the radiu

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$n = 2$

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(C) The radius of the $n^{th}$ orbit for a hydrogen-like atom is given by the formula:$r_n = r_0 \left( \frac{n^2}{Z} \right)$where $r_0$ is the radius of the first orbit of the hydrogen atom and $Z$ is the atomic number.For Beryllium $(Be^{+++})$,the atomic number $Z = 4$.We want the orbital radius $r_n$ to be equal to the ground state radius of hydrogen,which is $r_0$.Setting $r_n = r_0$,we get:$r_0 = r_0 \left( \frac{n^2}{4} \right)$$1 = \frac{n^2}{4}$$n^2 = 4$$n = 2$Therefore,the $n = 2$ state of triply ionised Beryllium has the same orbital radius as the ground state of hydrogen.

Which state of triply ionised Beryllium $(Be^{+++})$ has the same orbital radius as that of the ground state of hydrogen?

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