The acceleration of an electron due to the mutual attraction between the electron and a proton when they are 1.6 Å apart is, (me 9 × 10⁻³¹ kg, e = 1.6 × 10⁻¹⁹ C) (Take (1)/(4πε₀) = 9 × 10⁹ N m² C⁻²) (a) 10²⁵ m/s² (b) 10²⁴ m/s² (c) 10²³ m/s² (d) 10²² m/s²

Answer: (d). F = (ke²)/(r²) and a = F/me. a = 9×10⁹×(1.6×10⁻¹⁹)²9×10⁻³¹×(1.6×10⁻¹⁰)² = 9×10⁹×10⁻¹⁸9×10⁻³¹ = 10²² m/s² So the answer is (d) 10²² m/s².

The acceleration of an electron due to the mutual attraction between… | NEET 2020 (Phase 2)

Q62NEET 2020 · PHASE 2Easy

Electric Charge & Coulomb's Law

The acceleration of an electron due to the mutual attraction between the electron and a proton when they are 1.6 Å apart is,

$(m_e \simeq 9 \times 10^{-31}\text{ kg},\ e = 1.6 \times 10^{-19}\text{ C})$

(Take $\frac{1}{4\pi\epsilon_0} = 9 \times 10^{9}\text{ N m}^2\text{ C}^{-2}$ )

Show answer and solution

Answer: (d)

$F = \dfrac{ke^2}{r^2}$ and $a = \dfrac{F}{m_e}$ .
$a = \dfrac{9\times10^9\times(1.6\times10^{-19})^2}{9\times10^{-31}\times(1.6\times10^{-10})^2}$
$= \dfrac{9\times10^9\times10^{-18}}{9\times10^{-31}} = 10^{22}\ \text{m/s}^2$
So the answer is (d) $10^{22}\ \text{m/s}^2$ .