In hydrogen like system the ratio of coulombian force and gravitational force between an electron and a proton is in the order of :

Solution

Physics sir · 3 steps

Both forces fall off as 1/r², so the distance cancels: just compare ke² with Gmₑmₚ.

1. Distance cancels. Coulomb's law gives Fₑ = ke²/r² and Newton's law gives F_g = Gmₑmₚ/r². Both depend on 1/r², so the ratio is the same at every distance, including the orbit radius.

   Fₑ/F_g = ke²/(G mₑ mₚ)

2. Put in numbers. The numerator is 9 × 10⁹ × (1.6 × 10⁻¹⁹)² = 2.3 × 10⁻²⁸. The denominator is 6.67 × 10⁻¹¹ × 9.1 × 10⁻³¹ × 1.67 × 10⁻²⁷ = 1.01 × 10⁻⁶⁷, everything in SI units.
3. So. Fₑ/F_g = (2.3 × 10⁻²⁸)/(1.01 × 10⁻⁶⁷) ≈ 2.3 × 10³⁹, which is of the order of 10³⁹, option (b).

The idea to keep. For any pair of charged particles, the electric to gravitational force ratio does not depend on distance. For an electron and a proton it is about 10³⁹, so gravity is negligible inside atoms.

Answer from the NTA final answer key. Solution by PrepWiser.