Figure 54.5
-If plants in a northern temperate area use 26,400 kcal/m² of energy from solar radiation in photosynthesis,and the total amount of solar radiation energy reaching that area is 2,640,000 kcal/m²,what is the overall gross photosynthetic efficiency?
A)0)01%
B)0)1%
C)1%
D)100%

Question

Figure 54.5 
-If plants in a northern temperate area use 26,400 kcal/m² of energy from solar radiation in photosynthesis,and the total amount of solar radiation energy reaching that area is 2,640,000 kcal/m²,what is the overall gross photosynthetic efficiency?
A)0)01% 
B)0)1% 
C)1% 
D)100%

Quizplus · Accepted Answer

The overall gross photosynthetic efficiency can be calculated by dividing the amount of energy used in photosynthesis (26,400 kcal/m²) by the total amount of solar radiation energy available (2,640,000 kcal/m²) and then multiplying by 100 to convert the decimal to a percentage:

(26,400 / 2,640,000) x 100 = 1%

Therefore, the correct answer is C) 1%.

Figure 54.5 -If Plants in a Northern Temperate Area Use 26,400 Kcal/m²