Answer
See answer.
Work Step by Step
Assume that the uncertainty in the electron’s position is the nuclear radius. Use equation 28–1 to find the uncertainty in the electron’s momentum.
$$\Delta p=m\Delta v \geq \frac{\hbar}{\Delta x}$$
$$=\frac{(1.055\times10^{-34}J\cdot s)}{1\times10^{-15}m}= 1.055\times10^{-19}kg\cdot m/s$$
Changing units, this momentum is 198 MeV/c.
Assume that the lowest value for the electron momentum is this uncertainty. Now estimate the lowest possible energy. We must use a relativistic relation, equation 26-9. The rest mass of the electron is $mc^2 = 0.511 MeV$.
$$E=\sqrt{p^2c^2+m^2c^4}$$
$$E=\sqrt{(198MeV)^2+(0.511MeV)^2}\approx 200 MeV$$
The minimum energy is hundreds of MeV, which was to be proven.
