NEET | CLASS 12 | IMPORTANT QUESTIONS | PHYSICS | ELECTROSTATICS | The acceleration of an electron

The acceleration of an electron due to the mutual attraction between the electron and a proton when they are 1.6 Å apart is:
(Given: m = 9×10⁻³¹ kg, e = 1.6×10⁻¹⁹ C, (1 / 4πε₀) = 9×10⁹ Nm²C⁻²)

Options:
(a) 10²⁴ m/s²
(b) 10²³ m/s²
(c) 10²² m/s²
(d) 10²⁵ m/s²

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NEET, JEE Mains aspirants and Class 12 CBSE Physics students

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