Answer
See answers.
Work Step by Step
In Section 32–1, page 920, it is stated that the interaction energy for a collision in the LHC is about 14 TeV.
Example 32–1 shows us how to calculate the wavelength corresponding to this energy.
$$\lambda=\frac{hc}{E}=\frac{(6.626\times10^{-34}J \cdot s)(3.0\times10^8 m/s)}{(14\times10^{12}eV)(1.60\times10^{-19}J/eV)}=8.9\times10^{-20}m$$
This is approximately the level of detail that can be resolved by the collision of the proton beams.
We are asked to confirm that this is about 1/10000 the size of a nucleus. From equation 30–1, we know that a nuclear radius is about $1.2\times10^{-15}m$, so 1/10000 of that is $1.2\times10^{-19}m$.
This is indeed very comparable to the wavelength we just calculated.
