Question

The radiation dose used under Radappertisation process to prolong the shelf life of the product is

• A. 30 - 40 kGy
• B. 5 - 10 kGy
• C. 0.25 - 7.0 kGy
• D. 2.5 - 10 kGy

Hint:

Food Irradiation Doses:
Radappertization (Sterilization) = 30 - 40 kGy (High dose).
Radicidation (Pasteurization) = $2.5 - 10\text{ kGy}$.
Radurization (Extension) = $0.5 - 3.0\text{ kGy}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Seafood radiation preservation classifications: Radappertization achieves commercial sterilization (12D reduction of Clostridium botulinum spores) using high ionizing radiation doses of 30--40 kGy (30--50 kGy).
Key Formula or Approach:
\[ \text{Radappertization (Radiation Sterilization)} = 30 - 40\text{ kGy (30--50 kGy)} \quad [\text{Shelf-stable at Room Temperature}] \]

Step 2: Detailed Explanation:
Ionizing radiation preservation processes in food technology (IAEA/FAO/WHO standards):
1. Radappertization (Radiation Sterilization) (A): High-dose irradiation (30 to 40 kGy 30 to 50 kGy) applied to hermetically sealed pre-cooked foods. Achieves commercial sterility (12-decimal reduction of Clostridium botulinum spores), creating an indefinite shelf-stable product at ambient room temperature without refrigeration.
2. Radicidation (Radiation Pasteurization): Medium-dose irradiation ($2.5 - 10\text{ kGy}$) to kill specific non-spore-forming vegetative pathogens (e.g., Salmonella).
3. Radurization (Radiation Shelf-life Extension): Low-dose irradiation ($0.5 - 3.0\text{ kGy}$) to reduce spoilage psychrotrophs.

Step 3: Final Answer:
Thus, the dose under Radappertisation is 30 - 40 kGy, matching option (A).