A proton and neutron interact via the strong nuclear force. Their interaction is mediated by a particle called a meson, much like the interaction between charged particles is mediated by photons-the particles of the electromagnetic field. a) Perform a rough estimate of the mass of the meson from the uncertainty principle and the known dimensions of a nucleus \(\left(\sim 10^{-15} \mathrm{~m}\right)\). Assume the meson travels at relativistic speed. b) Use a line of reasoning similar to the one in part (a) to prove that the theoretically expected rest mass of the photon is zero. \(.\)

1. Using Heisenberg uncertainty principle: The Heisenberg uncertainty principle states that the product of the uncertainties in position (\(\Delta x\)) and momentum (\(\Delta p\)) of a particle must be at least equal to \(\hbar/2\). Mathematically, this can be expressed as: \[ \Delta x \Delta p \geq \frac{\hbar}{2} \] 2. Known dimensions of the nucleus: We are given the approximate dimensions of a nucleus as \(10^{-15}\ \mathrm{m}\). We can use this value as our uncertainty in position, \(\Delta x\). 3. Calculate the minimum uncertainty in momentum: Substitute the value of \(\Delta x\) in the Heisenberg uncertainty principle equation to get the minimum uncertainty in momentum, \(\Delta p\): \[ \Delta p \geq \frac{\hbar}{2 \Delta x} \] 4. Relativistic momentum equation: Since the meson is traveling at relativistic speeds, its momentum can be expressed in terms of its energy (\(E\)), mass (\(m\)), and speed of light (\(c\)) as: \[ p = \frac{E}{c} = \frac{mc}{\sqrt{1-\frac{v^2}{c^2}}} \] 5. Energy-momentum relation: For relativistic particles, the energy-momentum relation is given by: \[ E = \sqrt{(mc^2)^2 + (pc)^2} \] 6. Estimating the mass of the meson: Combining the minimum uncertainty in momentum obtained from the Heisenberg uncertainty principle, the relativistic momentum equation, and the energy-momentum relation, we can estimate the mass of the meson. \[ m \approx \frac{\hbar}{2\Delta x c} \]

1. Photons have zero mass: The rest mass of a photon is known to be zero, meaning \(m = 0\). We will now show that it is consistent with our previous reasoning. 2. Relativistic momentum equation for the photon: Since the photon has zero rest mass, its momentum is given by \(p = E/c\), where \(E\) is the energy of the photon. 3. Uncertainty principle for the photon: Considering the uncertainty principle for a photon, we have \(\Delta x \Delta p \geq \hbar/2\). 4. Relativistic limit of the position uncertainty: As the energy of the photon increases, its position uncertainty must decrease according to the uncertainty principle. In the limit case where \(E \rightarrow \infty\), the position uncertainty would be extremely small. 5. Conclusion: Since the photon has zero rest mass, its momentum is given by \(E/c\), and as the energy increases, the position uncertainty decreases, it is consistent with the photon having zero rest mass according to the uncertainty principle and the given line of reasoning.