The electron in a hydrogen atom has a kinetic energy of 13.6 eV in the ground state. What is the uncertainty in its position?

First, convert the energy to joules:

K = 13.6 eV * 1.60 x 10^-19 J/eV = 2.176 x 10^-18 J

Determine the magnitude of the electron's momentum:

K

=

mv2 2

=

m2v2 2m

=

p2 2m

p = (2mK)^1/2 = (2 * 9.11x10^-31 * 2.176 x 10^-18)^1/2

p = 1.99 x 10^-24 kg m/s

The electron's average momentum is actually zero, assuming the hydrogen atom is at rest. What we often do is assume that the magnitude of the electron's momentum is the same as the uncertainty in the momentum.

Therefore, Dp_y = 1.99 x 10^-24 kg m/s

Now use the uncertainty principle to find the uncertainty in the electron's position:

Dy

>

h 4p Dpy

=

6.63 x 10-34 4p * 1.99 x 10-24

= 0.5 x (5.30 x 10-11 m)

This happens to be half the radius of the ground-state electron's orbit in the Bohr model of the hydrogen atom.

What we've shown here is that if an electron has an energy typical for an electron in an atom, the position of the electron is uncertain and comparable to the size of the atom.