Question

Use the Heisenberg uncertainty principle to calculate Δx for an electron with Δv = 0.380 m/s.

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.)

Use the Heisenberg uncertainty principle to calculate Δx for a ball (mass = 108 g, diameter = 7.90 cm) with Δv = 0.380 m/s.

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

Answer 1

(a) here delta(V) = 0.380 m/s mass of electron m = 9.10 x 10^-31 Kg

as we know that from the heinsenberg principle

delta(x) x delta(P) <= h / (4 x pie )

delta(x) x m delta(V) = h / ( 4 x pie )

delta(x) x ( 9.1 x 10^-31 ) x 0.380 = ( 6.62 x 10^-34 ) / ( 4 x 3.14 )

delta(x) = 0.1524 x 10^-3 m Ans

(b) here the diometer of the hydrogen atom

d = 1.00 x 10^-8 cm = 1.00 x 10^-10 m

the factor is given by

( 0.1524 x 10^-3 ) / ( 1.00 x 10^-10 ) = 0.1524 x 10⁷

so the uncertaintity of electron position is large then tne diameter of the hydrogen atom

(c) if mass m = 108g = 0.108 Kg diameter = 7.90 cm = 7.90 x 10^-2 m delta(V) = 0,380 m/s

so the uncertaintity of position

delta(x) x 0.108 x 0.380 = ( 6.62 x 10^-34 ) / (4 x 3.14)

delta(x) = 12.84 x 10^-34 m Ans

(d) the factor of uncertaintity of position is given by

= (12.84 x 10^-34) / ( 7.90 x 10^-2 ) = 1.625 x 10^-32

it means when the factor of uncertaintity of position will 1.625 x 10^-32   then the uncertaintity of position is equal to the diameter of the atom