Question

A proton and a neutron interact via the strong nuclear force. Their interaction is mediated by 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 that the meson travels at relativistic speed. b) Use a line of reasoning similar to that in part (a) to prove that the theoretically expected rest mass of the photon is zero.

Short answer

Based on the Heisenberg Uncertainty Principle, the estimated mass of the meson that mediates the strong nuclear force between a proton and a neutron is approximately \(10^{-27} kg\). Also, the same principle proves that the theoretically expected rest mass of the photon is zero, confirming that photons are massless particles.

Step-by-step solution

Understanding the Heisenberg Uncertainty Principle

The Heisenberg Uncertainty Principle can be mathematically expressed as: \(\Delta x \cdot \Delta p \geq \frac{\hbar}{2}\), where \(\Delta x\) is the uncertainty in position, \(\Delta p\) is the uncertainty in momentum, and \(\hbar = \frac{h}{2\pi}\) is the reduced Planck constant.

Estimating the mass of the meson

As the size of the nucleus is given as approximately \(10^{-15} m\), we can set the uncertainty in position \(\Delta x\) equal to this value. Then, the uncertainty in momentum \(\Delta p\) can be expressed as: \(\Delta p \geq \frac{\hbar}{2 \Delta x}\). Now, we need to consider that the meson is traveling at relativistic speed. At these speeds, the relativistic momentum can be given by: \(p = \frac{mv}{\sqrt{1 - \frac{v^2}{c^2}}}\), where \(m\) is the mass of the meson, \(v\) is its velocity, and \(c\) is the speed of light. As \(v\) approaches \(c\), the denominator of the fraction approaches zero, meaning the momentum approaches infinity. Therefore, we can approximate the relativistic momentum as: \(p \approx m \cdot c\). Now, we can write the uncertainty in momentum as: \(\Delta p \approx \Delta m \cdot c\), where \(\Delta m\) is the uncertainty in the meson's mass. Combining this with the Heisenberg Uncertainty Principle, we get: \(\Delta m \cdot c \geq \frac{\hbar}{2 \Delta x}\). Now, we can solve for the mass of the meson \(\Delta m\): \(\Delta m \geq \frac{\hbar}{2 \Delta x \cdot c}\). Substituting the given values, we get: \(\Delta m \geq \frac{\hbar}{2 (10^{-15} \mathrm{m}) \cdot c} \approx 10^{-27} kg\). Thus, the estimated mass of the meson is approximately \(10^{-27} kg\).

Proving the theoretically expected rest mass of the photon

We know that photons have no rest mass (or they are massless). To prove this using the Heisenberg Uncertainty Principle, let's consider the case where the uncertainty in position becomes very large. In this case, the uncertainty in momentum will become very small, approaching zero. Since the photon travels at the speed of light, its momentum can be given as \(p = mc\). If the uncertainty in momentum approaches zero, this implies that the mass of the photon is also zero. Now, we can state our conclusion: based on the Heisenberg Uncertainty Principle, the theoretically expected rest mass of the photon is zero, confirming that photons are massless particles.