Question

Use the Heisenberg uncertainty principle to calculate ?x for an electron with ?v = 0.205 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 ?x for a ball (mass = 132 g, diameter = 6.40 cm) with ?v = 0.205 m/s.

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

Answer 1

heisenberg uncertainty principle:

mass *uncertainty in x*uncertainty in v >= h/(4*pi)

where h=planck’s constant

given uncertainty in v=0.205 m/s

then uncertainty in x >= (6.626*10^(-34)/(4*pi))/(9.1*10^(-31)*0.205)=2.8265*10^(-4) m

ratio of uncertainty in position and diameter of hydrogen atom

=2.8265*10^(-4)/(10^(-10) m)

=2.8265*10^(6) m

part B:

mass=132 gram=0.132 kg

then uncertainty in x >= (6.626*10^(-34)/(4*pi))/(0.132*0.205)=1.9486*10^(-33) m

ratio of uncertainty and diameter of the ball=1.9486*10^(-33)/(6.4*0.01 m)

=3.0447*10^(-32)