Question

1. Use the Heisenberg uncertainty principle to calculate Δx for an electron with Δv = 0.265 m/s.

2. By what factor is the uncertainty of the (above) electron's position larger than the diameter of the hydrogen atom? (Assume the diameter of the hydrogen atom is 1.00×10-8 cm.)

3. Use the Heisenberg uncertainty principle to calculate Δx for a ball (mass = 158 g, diameter = 6.55 cm) with Δv = 0.265 m/s.

4. The uncertainty of the (above) ball's position is equal to what factor times the diameter of the ball?

Answer 1

For minimum uncertainty, Heisenberg uncertainty principle is : ΔxΔp= h / 4π

where, Δp = mΔv, m = mass., h / 2π = 1.05 x 10^-34 J.sec

1. Δv = 0.265 m/s, m = 9.1 x 10^-31 kg.

Hence, Δx = 1.05 x 10^-34 / (2 x 9.1 x 10^-31 x 0.265) m = 2.18 x 10^-4 m = 0.218 mm

2. m = 158 gm = 0.158 kg, Diameter = d = 6.55 cm = 0.0655 m, v = 0.265 m/s

So, x = 1.05 x 10^-34 / (2 x 0.158x 0.265) m = 1.25 x 10^-33 m

3. The uncertainty of the (above) ball's position is (1.25 x 10^-33 / 0.0655) times or 1.9 x 10^-33 times the diameter of the ball.

answered by: anonymous