Answer
$ 7\times10^{-8}eV $
Work Step by Step
Find the minimum uncertainty in the energy by using equation 28–2.
$$\Delta E \geq \frac{\hbar}{\Delta t}$$
$$=\frac{1.055\times 10^{-34}J \cdot s}{1\times10^{-8}s}\frac{1eV}{1.60\times10^{-19}J}$$
$$=6.59\times10^{-8}eV \approx 7\times10^{-8}eV $$