Answer
$1.0\times10^{-15}\% $.
Work Step by Step
First find the uncertainty in the energy of the particle from Eq. 28–2.
$$\Delta E \geq \frac{\hbar}{\Delta t}$$
$$=\frac{1.055\times 10^{-34}J \cdot s}{12\times10^{-6}s}\frac{1eV}{1.60\times10^{-19}J}$$
$$=5.49\times10^{-11}eV $$
Now calculate the percentage.
$$\frac{\Delta E}{E}=\frac{5.49\times10^{-11}eV}{5.5\times10^6eV}\times 100=1.0\times10^{-15}\% $$
