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(B) The radius of the $n^{th}$ orbit for a hydrogen-like species is given by the formula: $r_n = 0.529 	imes \frac{n^2}{Z} \mathring{A}$.For the $1^{

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$n=2, Be^{3+}$

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(B) The radius of the $n^{th}$ orbit for a hydrogen-like species is given by the formula: $r_n = 0.529 	imes \frac{n^2}{Z} \mathring{A}$.For the $1^{st}$ Bohr orbit of a hydrogen atom $(n=1, Z=1)$: $r = 0.529 	imes \frac{1^2}{1} = 0.529 \mathring{A}$.Now, checking the options:For $Be^{3+}$ $(Z=4)$ with $n=2$: $r = 0.529 	imes \frac{2^2}{4} = 0.529 	imes \frac{4}{4} = 0.529 \mathring{A}$.Thus, the $2^{nd}$ orbit of $Be^{3+}$ has the same radius as the $1^{st}$ orbit of hydrogen.

Which hydrogen-like species will have the same radius as that of the $1^{st}$ Bohr orbit of a hydrogen atom?

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