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

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(B) The radius of an orbit in a hydrogen-like atom is given by the formula $r_n = 0.53 \frac{n^2}{Z} \mathring{A}$,where $n$ is the principal quantum

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$2$

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(B) The radius of an orbit in a hydrogen-like atom is given by the formula $r_n = 0.53 \frac{n^2}{Z} \mathring{A}$,where $n$ is the principal quantum number and $Z$ is the atomic number.For the ground state of hydrogen,$n = 1$ and $Z = 1$,so the radius is $r_H = 0.53 	imes \frac{1^2}{1} = 0.53 \mathring{A}$.For triply ionised beryllium $(Be^{3+})$,the atomic number $Z = 4$.We want the radius of the $n$-th state of $Be^{3+}$ to be equal to the ground state radius of hydrogen:$0.53 \frac{n^2}{4} = 0.53 	imes 1$$\frac{n^2}{4} = 1$$n^2 = 4$$n = 2$.Thus,the $n = 2$ state of $Be^{3+}$ has the same orbital radius as the ground state of hydrogen.

Which state of triply ionised beryllium $(Be^{3+})$ has the same orbital radius as that of the ground state of hydrogen?

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