Set 57 Problem #6

Set 57 Problem number 6

Problem

What is the approximate uncertainty in the velocity of a proton known to remain within a nucleus of diameter 3.4 * 10^-15 m? What kinetic energy would the proton (mass approximately 1.6 * 10^-27 kg) have at this velocity?

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 3.4 fermions = 3.4 * 10^-15 meter therefore implies an uncertainty of

momentum uncertainty: `dp = h / `dx = ( 6.62 * 10^-34 kg m^s / s ) / ( 3.4 * 10^-15 m) = 1.947 * 10^-19 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 = 1.947 * 10^-19 kg m/s / (1.6 * 10^-27 ks) = 1.216 * 10^8 m/s.

At this velocity the proton would have kinetic energy

proton KE: .5 m v^2 = .5 ( 1.6 * 10^-27 kg) * ( 1.216 * 10^8 m/s)^2 = 1.182 * 10^-11 J,

or about

( 1.182 * 10^-11 J) / ( 1.6 * 10^-19 J / ev) = 73.875 Mev.

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 ).