The radius of the 2^{nd} orbit of B^{4+} ion | vedclass

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Question

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

Vedclass · Accepted Answer

$0.4232 \ \mathring{A}$

Vedclass · Answer

(C) The radius of the $n^{th}$ orbit for a hydrogen-like species is given by the formula: $r = 0.529 	imes \frac{n^2}{Z} \ \mathring{A}$ Where $n$ is the principal quantum number (orbit number) and $Z$ is the atomic number. For the $B^{4+}$ ion: $n = 2$ $Z = 5$ (Atomic number of Boron) Substituting these values into the formula: $r = 0.529 	imes \frac{2^2}{5} \ \mathring{A}$ $r = 0.529 	imes \frac{4}{5} \ \mathring{A}$ $r = 0.529 	imes 0.8 \ \mathring{A}$ $r = 0.4232 \ \mathring{A}$

The radius of the $2^{nd}$ orbit of $B^{4+}$ ion is:

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