Answer
$d\approx4\times10^{-16}m$
Work Step by Step
Use equation 32–3 to estimate the range of the force, d.
$$mc^2\approx\frac{hc}{2 \pi d}$$
$$d\approx\frac{hc}{2 \pi mc^2}$$
$$d \approx \frac{(6.63\times10^{-34} J \cdot s)(3.00\times10^8m/s)}{2\pi (497.6\times10^{6}eV)(1.60\times10^{-19}J/eV)}$$
$$d\approx4\times10^{-16}m$$
