Yields

A beam of 50 $\mu$A on a 4 cm liquid hydrogen target gives a luminosity of 5.4$\times$10³⁷ which with a 1.6 msr solid angle means that the expected yields can be obtained by multiplying the cross sections in Table 1 by 8.6$\times10^{34}$. This would mean 2.4 counts/second or about 8600 counts/hour at the lowest yield point, E_e = 3.362 GeV, $\theta_p = 12.664^o$ and about 2.8 counts/second at the other Q² = 4.90 GeV² point. The cross sections are a factor of 4 or more higher at all of the other settings.