How much energy will be converted by the primary producer to net primary productivity from 20,000 J of available sunlight energy?

200 J

The correct answer is Option (1) → 200 J 

Incident sunlight

Total sunlight available = 20,000 J

Less than 50% of sunlight is PAR (Photosynthetically Active Radiation) .So, PAR ≈ 50% of 20,000 J = 10,000 J

Plants capture only 2–10% of PAR as Net Primary Productivity (NPP).

NPP calculation:

= 2% x 10,000

= \(\frac{2}{100}\) x 10,000 =200J