If an electron is revolving round the nucleus of He ion at a distance of 4.0 Å, the magnitude of cen — Atomic Structure Chemistry Question

💡 Solution & Explanation

Centripetal force equals electrostatic attraction $F = k \frac{Z e^2}{r^2}$. For He$^+$, $Z=2$. $F = (9 \times 10^9) \frac{2 \times (1.6 \times 10^{-19})^2}{(4.0 \times 10^{-10})^2} = 9 \times 10^9 \times \frac{2 \times 2.56 \times 10^{-38}}{16 \times 10^{-20}} = 9 \times 10^9 \times 0.32 \times 10^{-18} = 2.88 \times 10^{-9}$ N. Therefore, correct answer is A.