Q156 / 160 · AP EAMCET 2023 (15 May Shift 2) · Physics

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
    a $\alpha n^2$
  • B
    a $\alpha \frac{1}{n^2}$
  • C
    a $\alpha n^4$
  • Correct answer
    D
    $\alpha \frac{1}{n^4}$

Solution

Centripetal acceleration is given by $a=\frac{v^2}{r}$ We have $\begin{aligned} & \mathrm{v}=2.18 \times 10^6 \frac{\mathrm{Z}}{\mathrm{n}} \mathrm{m} / \mathrm{s} \\ & \mathrm{r}=\frac{0.53 \mathrm{n}^2 \times 10^{-10}}{\mathrm{z}} \mathrm{m} \\ & \mathrm{a} \propto \frac{1}{\mathrm{n}^2} \times \frac{1}{\mathrm{n}^2} \\ & \Rightarrow \mathrm{a} \propto \frac{1}{\mathrm{n}^4} \end{aligned}$

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

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