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  4. An electron is revolving around a nucleus at a distance 0.528 mathringA with velocity 2.18 × 10-6m s-1 .The acceleration of electron is

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Q. An electron is revolving around a nucleus at a distance $0.528\,\mathring{A}$ with velocity $ 2.18 \times 10^{-6}m\,s^{-1}$ .The acceleration of electron is

Motion in a Plane

Solution:

Using centripetal acceleration,
$a = \frac{\upsilon^2}{r}$ we get,
$a = \frac{(2.18 \times 10^6) (2.18 \times 10^6)}{0.528\times 10^{-10}} = 9 \times 10^{22} m \, s^{-1}$