Answer
1 MeV.
Work Step by Step
Assume that the uncertainty in the nucleon’s position is the nuclear radius. Use equation 30-1 to find the nuclear radius for iron.
Use equation 28–1 to find the uncertainty in the neutron’s momentum.
$$\Delta p=m\Delta v \geq \frac{\hbar}{\Delta x}$$
From that, find the uncertainty in its speed.
$$\Delta v \geq \frac{\hbar}{m\Delta x}$$
$$=\frac{(1.055\times10^{-34}J\cdot s)}{(1.67\times10^{-27}kg)56^{1/3}(1.2\times10^{-15}m)}= 1.38\times10^7m/s$$
The uncertainty in speed is about 5 percent of the speed of light. Using this as the nucleon's minimum speed, calculate its minimum kinetic energy. Use relativistic relations, though the answer is about the same if relativistic effects are ignored.
$$KE=mc^2(\frac{1}{\sqrt{1-v^2/c^2}}-1)$$
$$KE=939.6MeV(\frac{1}{\sqrt{1-0.0459^2}}-1)$$
Keep just 1 significant figure in this rough estimate.
$$KE\approx 1 MeV$$
