The centripetal acceleration ' $a$ ' of an electron in an orbit of hydrogen and the principal quantum number…
The centripetal acceleration ' $a$ ' of an electron in an orbit of hydrogen and the principal quantum number ' $n$ ' of the orbit are related by
- a $\alpha n^2$
- a $\alpha \frac{1}{n^2}$
- a $\alpha n^4$
- $\alpha \frac{1}{n^4}$
Solution
Centripetal acceleration is given by
$a=\frac{v^2}{r}$
We have
$\begin{aligned}
& \mathrm{v}=2.18 \times 10^6 \frac{\mathrm{Z}}{\mathrm{n}} \mathrm{m} / \mathrm{s} \\
& \mathrm{r}=\frac{0.53 \mathrm{n}^2 \times 10^{-10}}{\mathrm{z}} \mathrm{m} \\
& \mathrm{a} \propto \frac{1}{\mathrm{n}^2} \times \frac{1}{\mathrm{n}^2} \\
& \Rightarrow \mathrm{a} \propto \frac{1}{\mathrm{n}^4}
\end{aligned}$
Asked in: AP EAMCET 2023 (15 May Shift 2)
Practice more Structure of Atoms and Nuclei questions on Aicharya