Question #43829

A heavy nucleus is modeled as a 1D potential well along the $x$-axis (where $x$ is in $\text{m}$). The remaining nucleons are assumed to be fixed and act as the source of forces. To account for the saturation property of nuclear forces, the strong nuclear force $F_s$ on a single nucleon is modeled as zero in the core region $0 \le x \le \frac{R}{2}$. In the surface region $\frac{R}{2} < x \le R$, $F_s$ is a constant attractive force of magnitude $F_0$ (in $\text{N}$) directed towards the origin. Outside the nucleus ($x > R$), $F_s = 0$.

A neutron at rest at the origin ($x = 0$) undergoes beta decay, releasing a total energy $Q$ (in $\text{J}$). The decay produces an electron and an antineutrino which immediately escape, carrying away a combined kinetic energy $K_L$ (in $\text{J}$). The newly formed proton travels along the positive $x$-axis. Inside the nucleus ($0 \le x \le R$), the proton experiences the strong force $F_s$ and a constant electrostatic repulsive force of magnitude $F_e$ (in $\text{N}$) directed away from the origin.

Given that $F_0 > 2F_e$, and the proton just barely escapes the nuclear surface (reaches $x = R$ with zero velocity), what is the combined kinetic energy $K_L$ of the leptons?