An electron revolves in an excited orbit of hydrogen with speed v = √25.6 × 10⁵ m s⁻¹. Its orbit radius is x × 10⁻⁹ m. Find x. Take mₑ = 9 × 10⁻³¹ kg, |e| = 1.6 × 10⁻¹⁹ C and 1/(4πε₀) = 9 × 10⁹ N m² C⁻².
Options

Question

An electron revolves in an excited orbit of hydrogen with speed v = √25.6 × 10⁵ m s⁻¹. Its orbit radius is x × 10⁻⁹ m. Find x. Take mₑ = 9 × 10⁻³¹ kg, |e| = 1.6 × 10⁻¹⁹ C and 1/(4πε₀) = 9 × 10⁹ N m² C⁻².

Options

G1 Physics · Accepted Answer

Step 1: The electrostatic force supplies centripetal acceleration: mₑv²/r = ke²/r². Step 2: Rearrange for radius: r = ke²/(mₑv²). Step 3: Given v = √25.6 × 10⁵ m s⁻¹, so v² = 25.6 × 10¹⁰ = 2.56 × 10¹¹ m² s⁻². Step 4: r = [9 × 10⁹ × (1.6 × 10⁻¹⁹)²]/[9 × 10⁻³¹ × 2.56 × 10¹¹]. Step 5: Numerator = 2.304 × 10⁻²⁸; denominator = 2.304 × 10⁻¹⁹; r = 10⁻⁹ m. Final answer: x = 1 (Option D).