Q1 / 160 · AP EAMCET 2024 (21 May Shift 2) · Chemistry
The difference in radii between fourth and third Bohr orbits of $\mathrm{He}^{+}$(in m) is
- A$2.64 \times 10^{-10}$
- B$1.85 \times 10^{-12}$
- Correct answerC$1.85 \times 10^{-10}$
- D$1.85 \times 10^{-9}$
Solution
Bohr radii $\left(\mathrm{r}_{\mathrm{n}}\right)=52.9 \times \frac{\mathrm{n}^2}{\mathrm{Z}} \mathrm{pm}$
radii of fourth orbit of $\mathrm{He}^{+}=52.9 \times \frac{(4)^2}{2}$
radii of third orbit of $\mathrm{He}^{+}=52.9 \times \frac{(3)^2}{2}$
difference of radii $=52.9 \times \frac{(4)^2}{2}-52.9 \times \frac{(3)^2}{2}$ $\begin{aligned} & =\frac{52.9}{2} \times 7 \mathrm{pm} \\ & =185.15 \mathrm{pm} \\ & =185.15 \times 10^{-12} \mathrm{~m} \\ & =1.85 \times 10^{-10} \mathrm{~m} \end{aligned}$
radii of third orbit of $\mathrm{He}^{+}=52.9 \times \frac{(3)^2}{2}$
difference of radii $=52.9 \times \frac{(4)^2}{2}-52.9 \times \frac{(3)^2}{2}$ $\begin{aligned} & =\frac{52.9}{2} \times 7 \mathrm{pm} \\ & =185.15 \mathrm{pm} \\ & =185.15 \times 10^{-12} \mathrm{~m} \\ & =1.85 \times 10^{-10} \mathrm{~m} \end{aligned}$
Hence, the correct answer is option: 'C'.