Answer
14.8 MeV.
Work Step by Step
Assume that the uncertainty in the neutron’s position is the nuclear radius. 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.
$$\Delta v \geq \frac{\hbar}{m\Delta x}$$
$$=\frac{(1.055\times10^{-34}J\cdot s)}{(1.67\times10^{-27}kg)(1.2\times10^{-15}m)}= 5.3\times10^7m/s$$
The uncertainty in speed is about 18 percent of the speed of light. Using this as the neutron's minimum speed, calculate the neutron's minimum kinetic energy. We must use relativistic relations.
$$KE=mc^2(\frac{1}{\sqrt{1-v^2/c^2}}-1)$$
$$KE=939.6MeV(\frac{1}{\sqrt{1-0.1755^2}}-1)$$
$$KE=14.8 MeV$$
