THE ENERGETICS OF PHOTOSYNTHESIS 55 



Numerical example: 



I. I = 0.162 X 10"^ cal-cnv-sec-i 

 F = 14 cm- 

 t = 600 sec 



Ey = iFt = 0.136 cal 

 xo, = 15.4 jA 



U, = 0.005 X 15.4 = 0.077 ca! 

 rf = U,/Ei = 0.57 caj/cal 



TI. / = 0.327 X 10-^ cal-cm--.sec-i 

 F = 14 cm- 

 / = 600 sec 



£., = iFt = 0.275 cal 



xo, = 23.0 Ml 



U. = 0.005 X 23.0 = 0.115 cal 

 rj = U-i/E-i = 0.42 cal /cal 



From these two experiments we find the values of £■), E.2, U] and U,. Equa- 

 tion 24 can then be used to calculate h 



b = vo = liiTi j-p, = 0.71 cal cal 



Thus, Warburg and Negelein (30) found in their very first experiments in 

 photosynthesis that an average of 70% of the radiation energy absorbed could 

 be converted into chemical energy. 



§ 22 Application of the Quantum Theory 



In § 21 77 was expressed in cal/cal. If we express U in ^ul Oo, then r] and Vn 

 can be expressed in jul/cal ; then according to ecjuation 25 



1 ^1 O, = 0.005 cal 



1 mI cal = 0.005 cal/cal 



If U is expressed in mole Oo, then rj and rj^. would be expressed in mole/cal 

 whereby 



1 Ml/cal = 4.5 X 10-» mole/cal 



and 



1 mole/cal = 1.1 X 10^ cal/cal 



All these relations may be easily calculated from equation 25. 



In photosynthesis the value of rj is not constant for any particular wave- 

 length. Its value decreases as the wave-length decreases, as can be shown 

 from experiments made in red, yellow and blue light. 



Numerical example: 



a) Red light (6600 A) 



E = 0.208 cal 



.vo, = 24.2 m1 



U = 24.2 X 4.5 X 10-8 = 108.9 X lO"* mole 



r, = U/E = 108.9 X lO-VO-208 = 5.23 X lO-^ mole/cal 



