Set 57 Problem #5

Set 57 Problem number 5

Problem

What is the approximate uncertainty in the velocity of an electron (mass 9.11 * 10^-31 kg) known to remain within a distance of 1.4 Angstroms of a proton?

What kinetic energy would the electron have at this velocity?

Approximately how many times would the electron 'orbit' the proton in a circular 'orbit' at this distance and velocity?

Compare the centripetal acceleration at this distance and velocity to the acceleration of the electron due to the electrostatic force between an electron and a proton at this distance.

Solution

The product of the uncertainty in the position (standard units of meters) and the uncertainty in momentum (standard units of kg m/s) is equal to Planck's constant 6.63 * 10^-34 J s (units kg m^2 / s^2 * s = kg m^2 / s, same as units of position multiplied by units of momentum). We write this as

uncertainty principle: `dx * `dp = h.

The position uncertainty 2 * 1.4 Angstroms = 2.8 * 10^-10 meters, the diameter of the circle on which the electron 'orbits'. This implies an uncertainty of

momentum uncertainty: `dp = h / `dx = ( 6.63 * 10^-34 kg m^s / s ) / ( 2.8* 10^-10 m) = 4.735 * 10^-24 kg m/s.

Since momentum is the product p = m v of mass and velocity, the uncertainty in velocity is

velocity uncertainty: `dv = `dp / m = 4.735 * 10^-24 kg m/s / (9.11 * 10^-34 ks) = .5197 * 10^6 m/s.

At this velocity the electron would have kinetic energy

electron KE: .5 m v^2 = .5 ( 9.11 * 10^-31 kn) * ( .5197 * 10^6 m/s)^2 = 1.23 * 10^-19 J.

To find the time required to 'orbit' at the given distance, we will divide the circumference of the 'orbit' by the velocity:

time to orbit = circumference / velocity = 2 `pi * 1.4 * 10^-10 meters / ( .5197 * 10^6 m/s) = 16.91 * 10^-16 sec.

The frequency of the orbit is therefore

frequency = 1 / ( 16.91 * 10^-16 sec) = .05913 * 10^16 / sec.

The centripetal acceleration is

centripetal acceleration = v^2 / r = ( .5197 * 10^6 m/s) ^ 2 / ( 1.4 Angstroms) = ( .5197 * 10^6 m/s) ^ 2 / ( 1.4 * 10^-10 m ) = .1929 * 10^22 m/s^2.

The force on an electron with this acceleration is

centripetal force = m * centripetal acceleration = (9.11 * 10^-31 kg) * ( .1929 * 10^22 m/s^2) = 1.757 * 10^-9 N.

The magnitude of the force of attraction between proton and electron is

| Coulomb force | = | k q1 q2 / r^2 | = 9 * 10^9 N m^2 / C^2 * (1.6 * 10^-19 C) ^ 2 / ( 1.4 * 10^-10 m)^2 = 11.75 * 10^-9 Newtons.

Note the order-or-magnitude agreement between the result from the uncertainty principle and the result from Coulomb's Law.

General Solution

A particle confined within some distance `dx = 2 * r will have uncertainty `dp in its momentum, where `dx * `dp = h (Planck's constant).

Thus we have `dp = h / `dx = h / (2 * r ).

If we know the mass of the particle we can find from the momentum `dp = m `dv the velocity corresponding to a momentum equal to the uncertainty:

`dv = `dp / m = h / ( 2 * r * m )

The KE of the particle will be

KE = .5 m v^2 = ( m v ) ^ 2 / ( 2 m ).

An electron moving at this velocity in a circular orbit at distance r from a proton would require time

time to orbit proton: `dt = circumference / velocity = 2 `pi r / ( h / 2 * r * m ) = 4 `pi r^2 mE / h

to complete an orbit, where mE is the mass of an electron..

The centripetal force required to hold the electron in the orbit is

centripetal force = m v^2 / r = mE ( ( 4 `pi r^2 mE / h )^2 ) / r = 16 `pi^2 mE^3 r^3 / h^2.

This can be compared to the Coulomb attraction between electron and proton:

| Coulomb force | = k qE^2 / r^2,

where qE is the magnitude of the fundamental charge.