JEE Main 2024 Chemistry question
Based on Heisenberg's uncertainty principle, the uncertainty in the velocity of the electron to be found within an atomic nucleus of diameter 10⁻¹⁵ m is × 10⁹ ms⁻¹ (nearest integer) [Given: mass of electron = 9.1 × 10⁻³¹ kg, Plank's constant (h) = 6.626 × 10⁻³⁴ Js] (Value of π = 3.14)
Answer: 58
Solution
mΔvΔx = h/4π, so Δv = h/(4πmΔx) = 6.626 × 10⁻³⁴/(4 × 3.14 × 9.1 × 10⁻³¹ × 10⁻¹⁵) = 57.97 × 10⁹ m/s ≈ 58 × 10⁹ m/s.