Search any question & find its solution
Question:
Answered & Verified by Expert
The radius of the first orbit of $\mathrm{Li}^{2+}$ is $\mathrm{X} Å$. The radius of the third orbit of $\mathrm{He}^{+}$(in $Å$ ) is
Options:
Solution:
2717 Upvotes
Verified Answer
The correct answer is:
$\frac{27}{2} x$
Radius of 1st Bohr orbit, $\mathrm{r}=\frac{0.529 \mathrm{n}^2}{\mathrm{z}} Å$
$\begin{aligned} & \therefore \mathrm{r}_{\mathrm{He}^{+}}: \mathrm{r}_{\mathrm{Li}^{2+}}=\frac{0.529 \times 3^2}{2}: \frac{0.529 \times 1^2}{3}=\frac{9}{2}: \frac{1}{3} \\ & \therefore \mathrm{r}_{\mathrm{He}^{+}}=\frac{27}{2} x\end{aligned}$
$\begin{aligned} & \therefore \mathrm{r}_{\mathrm{He}^{+}}: \mathrm{r}_{\mathrm{Li}^{2+}}=\frac{0.529 \times 3^2}{2}: \frac{0.529 \times 1^2}{3}=\frac{9}{2}: \frac{1}{3} \\ & \therefore \mathrm{r}_{\mathrm{He}^{+}}=\frac{27}{2} x\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.