Question

In a salt marsh in Georgia, the quantity of solar radiation reaching the ground during the summer is 7,000,000cal/m2/day. The gross productivity of the producers in the salt marsh (mostly microscopic algae) is 0.0040 g/cm2/day. The average rate of respiration (the respiration requirement) is 35% of the gross primary productivity. Ecologists have found that 1 gram of salt marsh producers (the algae) is equivalent to 2,920 calories. Calculate the efficiency of photosynthesis.?

Answer 1

Answer:

$\mathtt{The \ efficiency \ of \ photosynthesis =1.08 \%}$

Explanation:

Given that:

the quantity of solar radiation = 7,000,000cal/m²/day

The producers' gross productivity in the salt marsh (mostly microscopic algae) is 0.0040 g/cm²/day.

= 0.004 × 2920 cal × 10000 m²/ day

= 116800 cal/m²/day

The respiration requirement is 35% of the gross primary productivity.

i.e  35% of 116800 calories/m²/day

=$\dfrac{ 35}{ 100 }\times 116800$ calories/m²/day

= 40880 calories/m²/day

The Net primary productivity  NPP = Gross productivity - Respiration  requirement

=( 116800 - 40880) calories/m²/day

= 75920 calories/m²/day

We can calculate the efficiency of photosynthesis by using the formula:

$\mathtt{The \ efficiency \ of \ photosynthesis = \dfrac{ Net \ primary \ productivity \ NPP}{quantity \ of \ solar \ radation} \times 100\%}$

$\mathtt{The \ efficiency \ of \ photosynthesis = \dfrac{ 75920 \ cal/m^2/day}{7000000 \ cal/m^2/day} \times 100\%}$

$\mathtt{The \ efficiency \ of \ photosynthesis =1.08 \%}$