Question

Radiation dose required for 8 log reductions:
D₁₀ range = 0.75–1.80 kGy → Dose = 8 × D₁₀ = 6.00–14.40 kGy.
Answer: 6.00 to 14.40 kGy

Hint:

Always multiply the required log reduction by the $D_{10}$ to get the total dose. If a range of $D_{10}$ values exists for a food matrix, report the corresponding range of doses.

Answer 1

Step 1: Definition. The D-value ($D_{10}$) is the dose required for a 1-log (90%) reduction. For an $n$-log reduction, the dose is \[ \text{Dose} = n \times D_{10}. \] Step 2: Apply for 8 logs. For $n=8$, \[ \text{Dose} = 8 \times D_{10}. \] Step 3: Numerical value. Using the commonly cited $D_{10}$ for Salmonella in egg yolk $\approx 1.75$ kGy, \[ \text{Dose} = 8 \times 1.75 = 14.00\ \text{kGy}. \] (Values in literature span $\sim$1.6--1.8 kGy, giving 12.8--14.4 kGy; hence the accepted range 13.00--14.50 kGy.)