An accelerated electron has a speed of 5 × 10⁶ ms⁻¹ with an uncertainty of 0.02%. The uncertainty in finding its location while in motion is x × 10⁻⁹ m. The value of x is ____________. (Nearest integer)
[Use mass of electron = 9.1 × 10⁻³¹ kg, h =6.63 × 10⁻³⁴ Js, π = 3.14]

Solution

Chemistry ma’am · 4 steps

Turn the 0.02% into an actual Δ v first, then Heisenberg gives Δ x in one line.

1. Find Δv. The uncertainty is 0.02% of the speed, so Δ v = 0.02/100 × 5 × 10⁶ = 1000 m s⁻¹.
2. Heisenberg principle. With Δ p = mΔ v, the principle Δ x · Δ p ≥ h/4π gives the minimum uncertainty in position.

   Δ x = h/(4π m Δ v) = (6.63 × 10⁻³⁴)/(4 × 3.14 × 9.1 × 10⁻³¹ × 1000)

3. Put in numbers. The denominator is 4 × 3.14 × 9.1 × 10⁻³¹ × 10³ = 1.143 × 10⁻²⁶. So Δ x = (6.63 × 10⁻³⁴)/(1.143 × 10⁻²⁶) = 5.80 × 10⁻⁸ m.
4. So. Δ x = 58.0 × 10⁻⁹ m, so x is 58.

The common slip: 116. Using h/2π in place of h/4π doubles the result and gives 116, not 58.

The idea to keep. A percentage uncertainty is about the speed, not about the momentum directly. Convert it to Δ v first, then use Δ x = h/(4π m Δ v), never h/2π.

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