The ratio of Coulomb's electrostatic force to the gravitational force between an electron and a proton separated by some distance is 2.4 × 1039. The ratio of the proportionality constant, K = (1)/(4πε₀) to the gravitational constant G is nearly (Given that the charge of the proton and electron each = 1.6 × 10 - 19 C, the mass of the electron = 9.11 × 10 - 31 kg, the mass of the proton = 1.67 × 10 - 27 kg): (a) 10 (b) 1020 (c) 1030 (d) 1040

Answer: (b). (Fe)/(Fg) = (Ke²)/(Gmemp) ⇒ K/G = (Fe)/(Fg)·(memp)/(e²) memp = 9.11×10⁻³¹×1.67×10⁻²⁷ ≈ 1.52×10⁻⁵⁷; e² = 2.56×10⁻³⁸ K/G = 2.4×10³⁹×1.52×10⁻⁵⁷2.56×10⁻³⁸ ≈ 1.4×10²⁰ So the answer is (b) 10²⁰.

The ratio of Coulomb's electrostatic force to the gravitational force… | NEET 2022 (Phase 2)

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The ratio of Coulomb's electrostatic force to the gravitational force between an electron and a proton separated by some distance is 2.4 $\times$ 1039. The ratio of the proportionality constant, $K = \frac{1}{4\pi\varepsilon_0}$ to the gravitational constant G is nearly (Given that the charge of the proton and electron each = 1.6 $\times$ 10 $-$ 19 C, the mass of the electron = 9.11 $\times$ 10 $-$ 31 kg, the mass of the proton = 1.67 $\times$ 10 $-$ 27 kg):

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Answer: (b)

$\frac{F_e}{F_g} = \frac{Ke^2}{Gm_em_p} \Rightarrow \frac{K}{G} = \frac{F_e}{F_g}\cdot\frac{m_em_p}{e^2}$
$m_em_p = 9.11\times10^{-31}\times1.67\times10^{-27} \approx 1.52\times10^{-57}$ ; $e^2 = 2.56\times10^{-38}$
$\frac{K}{G} = 2.4\times10^{39}\times\frac{1.52\times10^{-57}}{2.56\times10^{-38}} \approx 1.4\times10^{20}$
So the answer is (b) $10^{20}$ .

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