Q2 / 160 · AP EAMCET 2018 (23 Apr Shift 2) · Chemistry

If the ratio of energies of electron in the excited states of $\mathrm{H}$ and $\mathrm{Li}^{2+}$ is $1: 9$, the radius ratio of electron in the same excited states of $\mathrm{H}$ and $\mathrm{Li}^{2+}$ is

  • A
    $9: 1$
  • Correct answer
    B
    $3: 1$
  • C
    $1: 9$
  • D
    $1: 3$

Solution

$\because$ For same excited state $$ r \propto \frac{1}{Z} \text { and } E \propto Z^2 $$ Where, $r=$ radius of element $Z=$ atomic number of element ' $\mathrm{Z}$ ' for $(\mathrm{H})$ and $\left(\mathrm{Li}^{2+}\right)$ are 1 and 3 respectively thus, $$ \begin{aligned} & \frac{r_{(\mathrm{H})}}{r_{\left(\mathrm{L}^{2+}\right)}}=\frac{Z_{\left(\mathrm{Li}^{2+}\right)}}{Z(\mathrm{H})} \\ & \frac{r_{(\mathrm{H})}}{r\left(\mathrm{Li}^{2+}\right)}=\frac{3}{1}=3: 1 \end{aligned} $$

Hence, the correct answer is option: 'B'.

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