 Solution

Raising each beam energy to 250 MeV makes $E=500$ MeV. The kaon denominator is now dominated byrather than by $E^2{\Gamma }_K^2$. Substituting into a givesThe enormous change is a tiny-width effect, not a comparably abrupt change in the broad rho contribution. The center-of-momentum energy has moved about $2/(7.4\times 10^{-12})\simeq 2.7\times 10^{11}$ kaon widths from the peak. A width of $7.4\times 10^{-12}$ MeV is only $7.4\times 10^{-6}$ eV, requiring an energy-selection scale of order $10^{-14}$ relative to the several-hundred-MeV beam energy.

Real beams have vastly broader energy distributions. Averaging over such a distribution suppresses the useful narrow-resonance contribution: the finite peak value does not imply a large energy-integrated formation rate, whose resonant area shrinks with the width at fixed branching fractions. The weak inverse-decay entrance coupling and the broad strong-interaction rho background therefore make this a poor way to study kaon resonant formation. The supplied broad-resonance formula is used algebraically as requested; a practical calculation would convolve an appropriate narrow line shape with the actual beam distribution.