Q1 / 160 · TS EAMCET 2023 (12 May Shift 1) · Chemistry

The radius of third orbit of hydrogen atom is $\mathrm{R} \mathrm{pm}$. The radius of second orbit of $\mathrm{He}^{+}$ion (in pm is)

  • A
    $\frac{4}{3} R$
  • B
    $\frac{3}{4} R$
  • C
    $\frac{9}{2} R$
  • Correct answer
    D
    $\frac{2}{9} R$

Solution

$\begin{aligned} & \text { (d) Radius }=0.529 \times \frac{\mathrm{n}^2}{\mathrm{Z}} \\ & \text { for } 3^{\text {rd }} \text { orbit }(\text { of } \mathrm{H} \text { atom }) ; \mathrm{r}_3=\mathrm{R}=\frac{0.529 \times 9}{\mathrm{Z}} \\ & =4.761 \mathrm{pm}(\text { as } \mathrm{z}=1) \\ & \text { Now, for } 2^{\text {nd }} \text { wioit }\left(\text { of } \mathrm{He}^{+}\right) ; \mathrm{r}_2{ }^*=\frac{0.529 \times(2)^2}{2} \\ & =1.058 \mathrm{pm} \\ & \Rightarrow \mathrm{r}_2{ }^*=\frac{2}{9} \mathrm{r}_3 \text { or } \frac{2}{9} \mathrm{R} \mathrm{pm}\end{aligned}$

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

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