What is the radius ratio for the 2^{nd} orbit | vedclass

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(B) The radius of an orbit in a hydrogen-like species is given by the formula: $r_n = a_0 	imes \frac{n^2}{Z}$,where $a_0$ is the Bohr radius,$n$ is

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$16 : 27$

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(B) The radius of an orbit in a hydrogen-like species is given by the formula: $r_n = a_0 	imes \frac{n^2}{Z}$,where $a_0$ is the Bohr radius,$n$ is the orbit number,and $Z$ is the atomic number.For $Li^{+2}$ ion $(Z=3)$: $n_1 = 2$,$Z_1 = 3$. So,$r_1 = a_0 	imes \frac{2^2}{3} = a_0 	imes \frac{4}{3}$.For $Be^{+3}$ ion $(Z=4)$: $n_2 = 3$,$Z_2 = 4$. So,$r_2 = a_0 	imes \frac{3^2}{4} = a_0 	imes \frac{9}{4}$.The ratio is $\frac{r_1}{r_2} = \frac{4/3}{9/4} = \frac{4}{3} 	imes \frac{4}{9} = \frac{16}{27}$.

What is the radius ratio for the $2^{nd}$ orbit of $Li^{+2}$ ion to the $3^{rd}$ orbit of $Be^{+3}$ ion?

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