An electron is revolving in an excited state of a Hydrogen atom with velocity √(25.6)×10⁵ ms⁻¹. The radius of the orbit is x×10⁻⁹ m. The value of x is: [m_e=9×10⁻³¹ kg, e=-1.6×10⁻¹⁹ C, (1)/(4πε₀)=9×10⁹ Nm²C⁻²]
- A.4
- B.3
- C.2
- D.1✓
Correct Answer
(D) 1
Solution & Explanation
In a Bohr orbit the Coulomb attraction supplies the centripetal force: k e²/r² = m v²/r, where k = 1/(4πε₀) = 9×10⁹. Solving for r: r = k e²/(m v²). Given values: e = 1.6×10⁻¹⁹ C, m = 9×10⁻³¹ kg, v² = 25.6×10⁵ × ... note v = √25.6 ×10⁵, so v² = 25.6×10¹⁰ m²s⁻². Numerator: k e² = 9×10⁹ × (1.6×10⁻¹⁹)² = 9×10⁹ × 2.56×10⁻³⁸ = 2.304×10⁻²⁸. Denominator: m v² = 9×10⁻³¹ × 25.6×10¹⁰ = 230.4×10⁻²¹ = 2.304×10⁻¹⁹. r = (2.304×10⁻²⁸)/(2.304×10⁻¹⁹) = 1×10⁻⁹ m. So r = 1×10⁻⁹ m, giving x = 1, matching option D.
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