The radius of the $\mathrm{2^{nd}}$ orbit of $\mathrm{Li^{2+}}$ is $x$. The expected radius of the $\mathrm{3^{rd}}$ orbit of $\mathrm{Be^{3+}}$ is
Solution
$r_{\mathrm{Li}^{+}}=r_{0} \times \frac{2^{2}}{3}=x \Rightarrow r_{0}=\frac{3 x}{4}$
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$\mathrm{r}_{\mathrm{Be}^{3+}}=\mathrm{r}_{0} \times \frac{3^{2}}{4}$
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$r_{\mathrm{Be}^{3+}}=\frac{3 \mathrm{x}}{4} \times \frac{3^{2}}{4}=\frac{27 \mathrm{x}}{16}$
About this question
Subject: Chemistry · Chapter: Atomic Structure · Topic: Bohr's Model
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